nt.number theory - Why isn't there a structure with two primes?

I don't know whether this question is a bit too vague for MO or not, so feel free to delete it if you see fit.

The p-adic integer is defined by taking the inverse limit $ldots mathbb{Z} / p^2 rightarrow mathbb{Z}/p $.One way to see the p-adic integers is to see it as dealing with $ mathbb{Z} / p, mathbb{Z} / p^2, ldots $ at the same time. So $p$-adic integers allow us to see the structure of the ring of integers at the prime $p$. Taking the fractional field we obtain the $ p$-adic rational field $mathbb{Q}_p$.

This construction is useful in study the arithmetic of the ring (field). For example, in the theory of class field theory, we study the question in $mathbb{Q}_p$ first and glue them together and do something more to get the solution for $mathbb{Q}$.

I want to ask why it is not possible for us to construct a structure that will allow us to see the ring of integer at two primes $p,q$ together and see how do they interacts? The analog of the above inverse limit construction seems to still work though not an intergral domain. However, we can still localize where it is possible.

Here is the motivation for the above question. We know that the global question are not solve by simply gluing together solution for local question. I attribute that to the fact that the primes does not play alone but interact with one another. An illustration of this can be seen through the fact that the product of all normalized absolute value is 1. So my question is why not isolate two primes to understand how they are interacting with one another instead of looking at all of them at once. I think there is some complicated issue that will arise from this. Just want to know what they are.

A more particular question may be like this: Let call the construction obtained above $mathbb{Q}_{p,q} $. Is there something in the same vein of class field theory for this object. What is the obstacle in having such a theory. I am vaguely know that we have a more general Galois theory not only for fields but for rings also.

gn.general topology - Homotopy equivalences and cores

Hi all,

Before asking my question, I need to fix some terms and notation.

Let $M$, $M'$ be locally compact, Hausdorff spaces, and $f:Mrightarrow M'$ a homotopy equivalence with homotopy inverse $g:M'rightarrow M$. Set for a natural number $n$, $C^n(M):=(gcirc f)^n(M)$. Similarly, we set $C^{n}(M'):=(fcirc g)^n(M')$.

I'll call the data $(f,g,M,M')$ finitely-cored if there exist $N,N'in mathbb{N}$ such that $(gcirc f)$ restricted to $C^N(M)$ is the identity and $(fcirc g)$ restricted to $C^{N'}(M')$ is also the identity.

My questions are the following:

1) What are the properties of $(f,g,M,M')$ such that it is finitely-cored?

2) Under what conditions (if at all) can $(f,g)$ be deformed to another homotopy equivalence $(f_1,g_1)$ (i.e. $fsim f_1$, $gsim g_1$ )such that $(f_1,g_1,M,M')$ is finitely-cored?

3) For $(f,g,M,M')$ not finitely-cored, define $C_{infty}(M):= cap_{n=1}^infty C^n(M)$ and similarly for $M'$.
Is it true that $C_{infty}(M)$ is homeomorphic to $C_{infty}(M')$? If not, I'd also like to know under what conditions this holds.

Thanks in advance for your replies. Any references are also much appreciated!

P.S: I've invented the term "finitely-cored" here to make the question easier to present, so if there is already a term for this object I apologise for my ignorance.

soft question - The Importance of ZF

There is a very active ongoing debate within set theory about whether mathematics needs new axioms, and philosophers of mathematics are weighing in on all sides. Relevant considerations include many very deep topics in set theory, including independence, forcing and the large cardinal hierarchy. Some of these topics are at once highly technical and philosophical at the same time. It is fair to say that there is an emerging field called the philosophy of set theory that is grappling precisely with these issues.

Let me try to mention just a few of the considerations. First, the historical fact remains that the ZFC axioms are sufficiently powerful to carry out almost all of the construction methods that arise in mathematics outside set theory. Indeed, the ZFC axioms are provably far more powerful than necessary for the vast majority of ordinary mathematics. This is proved by the stunning results of the field of Reverse Mathematics (see Steve Simpson, Harvey Friedman etc.), which calculates for a huge collection of classical mathematical theorems exactly which axioms are needed to prove them. Reverse Mathematics proceeds by proving the axioms from the theorem as well as the theorem from the axioms (over a very weak base theory), thereby showing the necessity of those axioms, and it turns out that most all of the classical theorems of mathematics can be proved in relatively weak theories.

Nevertheless, within set theory, set theorists have discovered the ubiquitous independence phenomenon, by which an enormous number of set-theoretical assertions turn out to be independent of the ZFC axioms. This means that they are neither provable nor refutable in ZFC. We now have thousands of instances of fundamental set theoretic propositions that are known to be independent of ZFC. This includes almost any nontrivial statement of infinite cardinal arithmetic (such as the Continuum Hypothesis), as well as an enormous number of statements in infinite combinatorics, and so on. This phenomenon supports the view that ZFC is a weak theory, unable to decide these questions.

But of course, by the Incompleteness Theorem we know that any theory we can write down will exhibit this independence phenomenon. It is impossible in principle to avoid it.

Large cardinals are strong axioms of infinity, some of which go back to the time of Cantor (so they are not new), which are not provable in ZFC and which transcend ZFC in consistency strength, forming a vast hierarchy of consistency strength above it. Thus, they tend to make up for the weakness of ZFC (although there remains extensive independence even with large cardinals). Some set theorists make the case that the existence of large cardinals have numerous attractive regularity consequences, even for low down for sets of reals, that they seem to point the way towards the finally true set theory, which must remain elusively hidden from us because of the Incompleteness theorem. Making sense (or nonsense) of this view is a central concern of the emerging Philosophy of Set Theory.

Torsors in Algebraic Geometry?

I think I am confused about some terminology in algebraic geometry, specifically the meaning of the term "torsor". Suppose that I fix a scheme S. I want to work with torsors over S. Let $mu$ be a sheaf of abelian groups over S. Then my understanding is that a $mu$-torsor, what ever that is, should be classified by the cohomology gorup $H^1(X; mu) cong check H^1(X; mu)$.

Now suppose that $mu$ is representable in the category of schemes over S, i.e. there is a group object $$mathbb{G} to S$$
in the category of schemes over $S$, and maps (over S) to $mathbb{G}$ is the same as $mu$. Lots of interesting example arise this way.

I also thought that in this case a torsor over S can be defined as a scheme $P to S$ over S with an action of the group $mathbb{G}$ such that locally in S it is trivial. I.e. there exists a cover $U to S$ such that
$$ P times_S U cong mathbb{G} times_S U $$
as spaces over S with a $mathbb{G}$-action (or rather as spaces over U with a $mathbb{G} times_S U$-action).

The part that confuses me is that these two notions don't seem to agree. Here is an example that I think shows the difference. Let $S= mathbb{A}^1$ be the affine line (over a field k) and let $x_1$ and $x_2$ by two distinct points in $S$. Consider the subscheme $Y = x_1 cup x_2$, and let $C_Y$ be the complement of Y in S. Let $A$ be your favorite finite abelian group which we consider as a constant sheaf over S. Then we have an exact sequence of sheaves over S,
$$0 to A_{C_Y} to A to i_*A to 0$$
Where $i_*A(U) = A(U cap Y)$. I believe the first two are representable by schemes over S, namely $$C_Y times A cup S times 0$$
and $S times A$, where we are viewing the finite set $A$ as a scheme over $k$ (and these products are scheme-theoretic products of schemes over $spec ; k$).

In any event, the long exact sequence in sheaf cohomology shows that
$$H^1(S; A_{C_Y}) cong check H^1(S; A_{C_Y}) cong A$$
and it is easy to build an explicit C$check{text{e}}$ch cocycle using the covering given by the two opens consisting of the subschemes $U_i = S setminus x_i$, for $i = 1,2$.

Now the problem comes when I try to glue these together to get a representable object over S, i.e. a torsor in the second sense. Then I am looking at the coequalizer of
$$C_Y times A rightrightarrows (C_Ycup C_Y) times A$$
where the first map is the inclusion and the second is the usual inclusion together with addition by a given fixed element $a in A$. This seems to just gives back the trivial "torsor" $C_Y times A$.

Am I doing this calculation wrong, or is there really a difference between these two notions of torsor?

symplectic topology - Anosov diffeomorphisms and the chaotic hypothesis

Take the standard map with its break down of invariant circles and all,
Cantorii and the like. How would you start to approximate by Anosov?

I am thinking of Aubry/Mather/Bangert results on 2 degree of freedom
Hamiltonian systems, or, area preserving maps of an annulus to itself.
There are various theorems to the effect that these systems are a kind
of unentanglable mess of integrable and ``chaotic''.

On the flip side, there are results of Gole/Boyland asserting that IF
a natural mechanical system: kinetic + potential, admit a hyperbolic metric:
so the system is on $T^*Q$, and $Q$ admits a hyperbolic metric, then the system
is ``semi-conjugate'' to the corresponding Anosov system.

dg.differential geometry - Which Riemannian manifolds admit a finite dimensional transitive Lie group action?

At least in the compact case, there's a topological obstruction. In a 2005 paper, Mostow proved that a compact manifold that admits a transitive Lie group action must have nonnegative Euler characteristic. Here's the reference:

MR2174096 (2007e:22015)
Mostow, G. D.
A structure theorem for homogeneous spaces.
Geom. Dedicata 114 (2005), 87--102.

Of course, even if X admits a transitive Lie group action, most Riemannian metrics on X will not be homogeneous. (You didn't say whether you wanted actions by isometries, but I assume that's what you're interested in, because otherwise the Riemannian structure on X is irrelevant.) In the 2D case, the only compact, connected, homogeneous Riemann surfaces are the sphere, $RP^2$, the torus (and maybe the Klein bottle?)*, all with constant-curvature metrics. In general, the group has to be compact, because the isometry group of a compact Riemannian manifold is itself compact.

*EDIT 3: An earlier paper by Mostow constructed a transitive group action on the Klein bottle, but I doubt that this action preserves a Riemannian metric. It's a complicated construction, so I haven't had a chance to work through it in detail, but here's the reference:

Mostow, G.D., The Extensibility of Local Lie Groups of Transformations and Groups on Surfaces. Ann. Math., Second Series, (52) No. 3 (1950), 606-636.

I don't know what's known in the noncompact case.

ca.analysis and odes - Existence of an "anti-additive" (or "never linear") map?

(I've edited this question)

I'm searching for a continuously differentiable function $f:mathbb R^2tomathbb R$ such that $f(x)+f(x+u+v)neq f(x+u)+f(x+v)$ for all $x$ and all linearly independent $u$ and $v$.

My original question was about the special case $x=0, f(x)=0$ for merely continuous functions, which turned out to be trivial.

(I was lead to this question when investigating whether one can always find the vertices of a parallelogram (or more specifically, a square) in the graph of a continuously differentiable function $f:mathbb R^2tomathbb R$. The nonexistence of functions such as the above would imply that one cannot always find a parallelogram in the graph of a continuously differentiable function.)

Infinite products of topological groups

While studying for a topological groups course, I wondered if we could define the product of uncountably many topological groups such that the product is still a topological group. That is: let $G_i$ be a topological group with product law $p_i$ for each $i in I$ (with $I$ uncountable). We can give $G = prod_{i in I} G_i$ the (Tychonoff) product topology and define the product law of $G$ by:

$pi_i circ p = p_i$ for all $i in I$.

However, when trying to prove that this mapping is continuous end up needing $I$ to be at most countable or that the topologies of $G_i$ be discrete.

Is there any way to get around this?

Thanks.

rt.representation theory - Is the category of representations of a finite W-algebra monoidal?

This is just to add another point of view to Ben's and David's comments and mainly for your Edit regarding superconformal field theories, as this is more about the affine case than the finite one.

1) There is a formal relation between the finite W-algebra and the affine one: The finite is the Zhu algebra of the affine one (Kac-De Sole '05).

This in particular implies that irreducible modules for the affine algebra are in 1-1 correspondence with irreducible modules for the finite one. From here perhaps your suspicion that if you expect tensor products for modules of the affine ones then you should expect them for the finite one. That article also answer your question regarding a BRST construction of the finite W-algebra (see the appendix, and also the very nice article by Gan-Ginzburg in the finite case)

2) There is a fusion category structure for the affine W-algebra at certain levels, namely, it is proved in some cases when these W-algebras are rational, and it is still conjectured in many others. At any rate, we know of at least a few examples when we indeed have the tensor structure in the categories of modules for the affine W-algebra (See articles by Arakawa and Kac-Wakimoto starting in 2005 up to late '10).

One observation is that the affine W-algebra that is rational is the simple quotient of the affine W-algebra mentioned in 1) above.

3) The functor that you mention is just the "top component" of the homological reduction functor in the affine situation, where you feed a representation for the affine algebra g and you get a representation for the affine W-algebra (the simple one). This functor has been studied by many, there was a conjecture of Frenkel-Kac-Wakimoto regarding the exactness and behaviour of this functor. Arakawa proved (for the principal nilpotent first in Invent. '05 and then in more generality starting in '08, see also Kac-Wakimoto '07) that this functor is exact and it sends irreducible modules to either zero or irreducible modules. He used this to prove existence of modular invariant representations of the W-algebra. In particular, as you mention this gives a way of fusing W-representations by using the fusion of the affine ones.

Actually, Arakawa proved the irreducibility part in the principal nilpotent case, and as far as I know the almost-irreducibility in general (0802.1564). In some cases he can use a previous result in the finite case: this property of sending irreducibles to irreducibles or zero was proved by Brundan-Kleschev in type A (they proved more: that every simple module over W-finite arises in this way).

4) In the cases when you can prove that a) the affine W-algebra is rational, and b) that every module over the affine W-algebra is the Hamiltonian reduction of a module over the affine lie algebra g, then you get fusion for the W-algebra from fusion on g, and I do not see anything wrong with passing to the finite W-algebra by taking Zhu's algebras everywhere, so in this setting I agree with you, and looking at the list in Kac-Wakimoto '07 of possible rational W-algebras, you should get several examples of finite ones. I don't see anything wrong with this but I may be missing something, perhaps some subtlety between the W-algebra and its simple quotient?

ag.algebraic geometry - equivalence of Grothendieck-style versus Cech-style sheaf cohomology

Let $X$ be a topological space, and $T$ its category of open sets with the usual Grothendieck topology. Let $T'$ be any sieve of $T$ (a subcategory of $T$ such that if $U$ is in $T'$ then any subset of $U$ is also in $T'$). For example, $T'$ might be the collection of open subsets subordinate to the open subsets in a cover $mathcal{U}$. Any sheaf on $T$ induces a functor on $T'$ which can be viewed as a sheaf on $T'$ if $T'$ is given the minimal topology (the only covers are the identity maps). This determines a morphism of topoi $f : T rightarrow T'$, hence a spectral sequence

$H^p(T', R^q f_ast F) Rightarrow H^{p+q}(T, F)$ .

(One could surely also convince oneself that such a spectral sequence exists without any reference to topoi.)

The Cech cohomology of $F$ with respect to some covering family $mathcal{U}$ is

$H^p(mathcal{U}, F) = H^p(T', f_ast F)$

where $T' = T'(U)$ is the sieve associated to the cover $mathcal{U}$. The Cech cohomology is then the filtered colimit

$check{H}^p(T, F) = varinjlim_{(T',f)} H^p(T', f_ast F)$

taken over the projections $f : T rightarrow T'$ associated as above to covering families $mathcal{U}$.

One evidently has edge homomorphisms

$check{H}^p(T, F) rightarrow H^p(T, F)$

from the spectral sequence, and the question is when these induce an isomorphism. If we could somehow eliminate the $R^p f_ast F$, $p > 0$, by passing to a "small enough" cover we would have equality. This condition already holds in many cases; the following condition is more general (but I haven't checked carefully that it actually works!):

For every cover $mathcal{U}$ of $X$, every $U_1, ldots, U_n in mathcal{U}$, and every class in $alpha in H^p(U_1 mathop{times}_X cdots mathop{times}_X U_n, F)$, $p > 0$, there exists a refinement $mathcal{U}'$ of $mathcal{U}$ such that the restriction of $alpha$ under the map

$H^p(U_1 mathop{times}_X cdots mathop{times}_X U_n, F) rightarrow H^p(U'_1 mathop{times}_X cdots mathop{times}_X U'_n, F)$

is zero.

To make sense of this, one must use some convention for the covers $mathcal{U}$ and $mathcal{U}'$ to ensure there is a map as above. For example, one could work only with covers indexed by the points of $X$ (a cover is then a collection of neighborhoods of each point of $X$).

A more refined version of the above condition would say that Cech cohomology equals cohomology in degrees at most $q$ if the above condition holds for $p leq q$. Since it always holds for $p = 0,1$ this implies that

$check{H}^1(T, F) = H^1(T, F)$

in general.

Edit in response to David's comment:

The Cech complex always computes cohomology correctly in a presheaf category (i.e., when the topology is "chaotic": an object has no covers by anything except itself). Trying to compute cohomology in an arbitrary site using the Cech complex is (heuristically) something like trying to approximate the site by a presheaf category.

Here is how Cech cohomology computes cohomology of presheaves. Consider any category $T'$. If $F$ is a presheaf of groups on $T'$ then the sheaf cohomology groups of $F$ are the derived functors of the inverse limit for diagrams of shape $T'$. They are also computed as

$Ext(mathbf{Z}, F)$

where $mathbf{Z}$ is the constant sheaf associated to the integers. Remarkably, in a presheaf category, $mathbf{Z}$ has a canonical projective resolution associated to any cover of the final presheaf. A cover of the final presheaf is a collection of objects $U$ of $T'$ such that every object of $T'$ has a map to at least one object of $U$. The $i$-th term of this complex is the direct sum, over all choices of $i$ elements $U_1, ..., U_i$ of $U$, of the groups $mathbf{Z}_{U_1 times cdots times U_i}$. (You can check this is projective by noting it is the extension by $0$ of $mathbf{Z}$ from the slice category $T' / U_1 times cdots times U_i$ and extension by $0$ preserves projectives (since it has an exact right adjoint) and $mathbf{Z}$ is projective on the slice category since all higher cohomology of all sheaves vanishes (since it has a final object). It's also easy to check by a direct calculation.)

Denote this complex by $K$. Since this is a projective resolution of $mathbf{Z}$, $mathrm{Hom}(K, F)$ computes the cohomology of $F$. But it is also easy to see that this is just the Cech complex of $F$.

ct.category theory - Infinity groupoid objects

There is not a model category on simplicial smooth manifolds which gives the data you're looking for, but there is a structure of category of fibrant objects such that nerves of groupoids are fibrant. The fibrations are Kan maps (a la Henriques, but with the additional requirement that the maps on vertices be submersions), and the weak equivalences are maps which induce isomorphisms on all simplicial homotopy groups (again, as Henriques defines them). Nerves of groupoids are fibrant in this sense, hypercovers are trivial fibrations, and morphisms qua principal bibundles are equivalent to morphisms via spans where the source leg is a hypercover.

This category of fibrant objects structure doesn't extend to a model structure on simplicial manifolds because there aren't any positive dimensional cofibrant objects (e.g. hypercovers are trivial fibrations, but given any positive dimensional manifold, you can build a hypercover on it which doesn't admit a section).

If you still want to work in a model category capturing this, the natural thing is to use Yoneda and pass to the local model structure on simplicial presheaves. You can show that this is a fully faithful embedding on the level of homotopy categories.

algorithms - Other norms for Lattice reduction techniques (LLL, PSLQ)?

LLL and other lattice reduction techniques (such as PSLQ) try to find a short basis vector relative to the 2-norm, i.e. for a given basis that has $ varepsilon $ as its shortest vector, $ varepsilon in mathbb{Z}^n $, find a short vector s.t. $ b in mathbb{Z}^n, ||b||_2 < ||c^n varepsilon||_2 $.

Has there been any work done to find short vectors based on other, potentially higher, norms? Is this a meaningful question?

homological algebra - Why does non-abelian group cohomology exist?

I don't know if this is a "deep" or "shallow" explanation, but if anyone is still reading this thread, here is a different explanation. I'll start with the preliminary comment that the cohomology of a group $K$ is a special case of cohomology of topological spaces. In topology in general, you get the same phenomenon that $H^k(X,G)$ is well-defined either when $G$ is abelian or $k=1$.

Consider the definition of simplicial cohomology for locally finite simplicial complexes. Or, more generally, CW cohomology for locally finite, regular complexes — regular means mainly that each attaching map is embedded. You can define a $k$-cochain with coefficients in a group $G$ (or even in any set) as a function from the $k$-cells to $G$. In attempting to define the coboundary of a cochain $c$ on a $k+1$-cell $e$, you should multiply together the values of $c$ on the facets of $e$. The obvious problem is that if $G$ is non-abelian, the product is order-dependent. However, if $k=1$, geometry gives you a gift: The facets are cyclically ordered, and what you mainly wanted to know is whether the product is trivial. The criterion of whether a cyclic word is trivial is well-defined in any group, not just abelian groups. A similar but simpler phenomenon occurs for the notion of a coboundary: If $e$ is an oriented edge and $c$ is a 0-cochain, there is a non-abelian version of modifying a 1-cochain by $c$ because the vertices of $e$ are an ordered pair.

So far this is just a more geometric version of Eric Wofsey's answer. It is very close to the fact that $pi_1$ is non-abelian while higher homotopy groups are abelian — and therefore non-commutative classifying spaces exist only for $K(G,1)$. However, in this version of the explanation, something extra appears when $X=M$ is a 3-manifold.

If $M$ is a 3-manifold, then not only are the edges of a face cyclically ordered, the faces incident to an edge are also cyclically ordered. It turns out that, at least at the level of computing the cardinality of $H^1(M,G)$, you can let $G$ be both non-commutative and non-cocommutative. In other words, $G$ can be replaced by a finite-dimensional Hopf algebra $H$ which does not need to be commutative or cocommutative. Finiteness is necessary because it is a counting invariant. The resulting invariant $\#(M,H)$ was a topic of my PhD thesis and is explained here and here. Although the motivation is original, the invariant is a special case of more standard quantum invariants defined by other people. (The same construction was also later found by three physicists, but I can't remember their names at all right now.)

Many 3-manifolds are also classifying spaces of groups, so for these groups there is the same notion of noncommutative, non-cocommutative group cohomology.

at.algebraic topology - Is any interesting question about a group G decidable from a presentation of G?

I don't have a complete answer, but here are some thoughts.

The Rips Construction takes an arbitrary finitely presented group Q and produces a 2-dimensional hyperbolic group $Gamma$ and a short exact sequence

$1to Kto Gammastackrel{q}{to} Qto 1$

such that the kernel $K$ is generated by 2 elements. It turns out, by a result of Bieri, that $K$ is finitely presentable if and only if $Q$ is finite.

One can improve the finiteness properties of $K$ using a fibre product construction. Let

$P={(gamma,delta)inGammatimesGammamid q(gamma)=q(delta)}$.

By the '1-2-3 Theorem', if $Q$ is of type $F_3$ then $P$ is finitely presentable.

I would guess that $P$ has good higher finiteness properties if and only if $Q$ is finite. Perhaps one can use the fact that $Pcong K rtimesGamma$.

Even if this is true then it still doesn't quite solve your problem, as we don't have a presentation for $P$. To do this, one needs to be given a set of generators for $pi_2$ of the presentation complex of $Q$, which enable one to apply an effective version of the 1-2-3 Theorem. (In the absence of this data, presentations for $P$ are not computable. Indeed, $H_1(P)$ is not computable.)

Question: Does there exist a list of presentations for groups $Q_n$ such that:

1. each group $Q_n$ is of type $F_3$;




2. the set ${ninmathbf{N}mid Q_ncong 1}$ is recursively enumerable but not recursive;




3. but generators for $pi_2(Q_n)$ (as a $Q_n$-module) are computable?

If so, and if I'm right that the higher finiteness properties of $P$ are determined by $Q$, then higher finiteness properties are indeed undecidable. Simply apply the Rips Construction and the effective version of the 1-2-3 Theorem to the list $Q_n$.

algorithms - Paritioning a set of numbers A into two sets B,C so that abs(prod(B) - prod(C)) is minimal

I suspect that it's NP-hard even to check whether you can get prod(B) - prod(C) = 0, although there's a problem with the obvious argument that I don't know off the top of my head how to fix.

"Reduction" from subset sum: If you have a set S of integers, replace each integer $k in S$ with $2^k$. Then this new set can be partitioned into two parts with the same product iff the original set could be partitioned into two parts with the same sum.

The problem is that our new integers are exponentially large compared to the original ones, which means that this isn't actually allowed as a reduction. But I think it's morally correct, since the hardness of subset sum is controlled by the size of the set rather than the lengths of the elements.

at.algebraic topology - Topologists loops versus algebraists loops

This is not an answer, but these are my thoughts so far and hopefully they will lead someone to a correct answer (hence the community wiki). My vague recollection is that the algebraic loop space only sees stuff "near" the constant loops, which is consistent with moonface's comments. I apologize if I am missunderstanding anything (I'm one of the struggling topologists). Basically I want to look at an example, which I think will elucidate the matter.

Let's take $X = mathbb{G}_m$, the multiplicative group. Then $ mathbb{G}_m(mathbb{C}) = Spec mathbb{C}[b, b^{-1}]$. As an analytic space I think this is just $mathbb{C}^times$, so on the topological side we get an interesting loop space. We have a fibration sequence,

$$Omega mathbb{C}^times to Lmathbb{C}^times to mathbb{C}^times $$

and since topologically $mathbb{C}^times simeq S^1$, this shows that $pi_0(L mathbb{C}^times) cong mathbb{Z}$. This is something that we should be able to detect if the algebraic version of the loop space is similar to the topological one, just count the number of components.

So what is the algebraic loop space in this case? Well, I guess by definition it is $Spec ; mathbb{C}((t))[b,b^{-1}]$. Now remind me, how do we turn this into a space? and how many components does it have?

If you try to take the $mathbb{C}$-points of it, i.e. homomorphisms,
$$ mathbb{C}((t))[b, b^{-1}] to mathbb{C}$$
don't you just get $mathbb{C}^times$? This seems to suggest that it is an infinitesimal thickening of the constant loops.

ag.algebraic geometry - Normal bundle of $CP^n$ in $CP^{n+1}$

Let $S^1$ act on $S^{2n+1}$ via Hopf action and $S^1$ also acts on $mathbb{R}^2$ via rotation about the origin.
Then $S^1$ acts on $S^{2n+1}times mathbb{R}^2$ diagonally.

Let $M$ be the quotient of this diagonal action.

My question is why $ M$ can be viewed as the normal bundle of $mathbb {CP}^n$ in $mathbb {CP}^{n+1}$.

I have a feeling that it must be related to the fact that: after removing a $(2n+2)$ disk in $mathbb {CP}^{n+1}$, the boundary $S^{2n+1}$ is fibered over the $mathbb {CP}^n$.

But where can I find the proof of the statement.

Integrable solutions to an elliptic PDE on divergence form have a definite sign

I love this problem and have spent half the evening thinking about it.

Here is a rough sketch of an idea that could possibly work. Making it rigorous might be a bit of a chore due to the unboundedness of $mathbb{R}^n$, etc, but my guess is that it is doable.

Let $L$ denote the elliptic operator [ Lu : = -Delta u - mathrm{div}( f u ). ]

Suppose the operator $L$ has a principal eigenvalue $lambda_1$, which is the smallest number $lambda$ for which there exists a nonzero solution of the equation $Lu = lambda u$ in $mathbb{R}^n$. Then $lambda_1$ should be simple (!) and have a principal eigenfunction which does not change sign. Let $varphi in L^1(mathbb{R}^n)$ denote the principal eigenfunction, which we normalize to be positive.

Assume for now that $varphi$ and its derivatives tend to zero at infinity, and $f$ and its derivatives stay bounded. Then we may simply integrate the equation $Lvarphi = lambda_1varphi$ to get $lambda_1 int_{mathbb{R}^n} varphi dx = 0$. Well, that means that $lambda_1 = 0$.

Recalling the simplicity of $lambda_1=0$, we see that the equation $Lu = 0$ not only has a positive solution, its set of solutions is precisely ${ c varphi : cin mathbb{R} }$. This implies the result.

Now, you may have already noticed that there aren't really such principal eigenvalues in general, for example when $f=0$. But I think it is possible that the idea can still be made into a rigorous proof.

If we look at a really large domain, say the ball $B(0,R)$ with $R> 0$ very large, there is a principal eigenvalue $lambda_{1,R}$ of $L$ on $B(0,R)$ and it is going to be close to zero. This can be shown by considering the adjoint operator $L^*$, which has the same principal eigenvalue $lambda_{1,R}$, and it is easy to show that $0 < lambda_{1,R} leq CR^{-2}$, due to boundedness of $f$. This is where we use the special form of the equation.

If we have a solution to $Lu = 0$ on the whole space $mathbb{R}^n$, with $u(0)>0$, I believe it should be possible to prove that if we choose the normalization $varphi_{1,R}(0) = u(0)$, then $varphi_{1,R}$ converges to $u$ (at least locally uniformly) as $Rto infty$. In particular, $u$ must be positive everywhere.

There are obviously some details left to work out. It is very possible I am making a silly error and it doesn't work at all.

computability theory - Differentiability of computable functions

You may be interested in some very recent work by Brattka, Miller and Nies looking at points of differentiability for computable functions in terms of algorithmic randomness. Briefly call a real x computably random (Martin-Löf random) if no computable (computably enumerable) martingale succeeds on a binary representation of x. Brattka, Miller and Nies show that:

1) At each computably random real, every computable function that is non-decreasing is differentiable.

2) At each Martin-Löf random real, every computable function of bounded variation is differentiable.

gr.group theory - Naturally occuring groups with cardinality greater than the reals.

I would expect that automorphism groups of natural structures would count as natural groups in your sense. But automorphism groups of uncountable structures often have size larger than the continuum. In general, the size of the automorphism group of a structure of size $kappa$ is bounded above by $2^kappa$, which is strictly larger than $kappa$, and this upper bound is often reached, when the structure is insufficient to restrict the general nature of automorphisms. For example, the number of bijections of an infinite set of size $kappa$ with itself is $2^kappa$.

I am sure that you will be able to construct many other natural structures of uncountable size $kappa$, whose automorphism groups have size $2^kappa$, and these would seem to the sort of examples you seek.

---

P.S. Let me also note that your remark that the reals have size $aleph_1$ is only correct when the Continuum Hypothesis holds. In general, the size of the reals, also known as the continuum, is $2^{aleph_0}$, which is also denoted $beth_1$, whereas $aleph_1$ is simply the first uncountable cardinal.

How much of differential geometry can be developed entirely without atlases?

This is a comment, not an answer, but is too long to fit in the comment box: having read the question, answers, and comments, I don't quite follow the intent of this question:

We can define a manifold to be a locally ringed space in which each point has a neighbourhood
isomorphic to an open subset of ${mathbb R}^n$ (or even just ${mathbb R}^n$ itself) with its sheaf of smooth functions (plus second countability and Hausdorffness, if you like). As was remarked by Dmitri, the collection of all such will then form an atlas, but one doesn't need to say this.

As Pete Clark says, what I've said so far is evident.

But it seems that another aspect of the question is whether one can always avoid working in coordinates. This seems to have nothing to do with atlases.

E.g. in arguments in the locally ringed space set-up, one will certainly in many instances verify that a property can be checked locally, and then verify it on Euclidean space with
its natural smooth structure. (Just as in the theory of schemes, one often shows that a property is local, and then checks it in the affine case.)

Now one can ask: can one avoid the latter kinds of arguments? This seems unlikely: manifolds are defined to be locally Euclidean (no matter which of the possible formalisms one is using), and so if one is proving theorems about manifolds, one will
have to use this somewhere. For example, one can surely define the tangent sheaf in
a coordiante free way, but to prove that it is locally free of rank equal to the dimension
of the manifold, one is going to reduce to a local computation and then appeal to calculus
on Euclidean spaces; there is no other way!

[EDIT: The last sentence may be too categorical of a declaration; see Dmitri Pavlov's answer for a suggestion of a more substantially algebraic reformulation of the
notion of manifold.]

at.algebraic topology - cell complexes and higher graph theory

Suppose that, on an intuitive basis, one defines a "2-graph" $(V,E,F,partial)$ as a collection of vertices, oriented edges and oriented faces, all of which should be considered as abstract objects whose relationships are determined by an incidence relation $partial$ such that

• $partial_0$ is the usual incidence matrix of graph $(V,E)$


• $partial_1$ is defined in an obvious way as the edge-face incidence matrix with the only requirement that the boundary of a face be an oriented cycle, so that $partial^2 = 0$.

Let a realization of a 2-graph be an embedding of vertices, lines and surfaces in $mathbb{R}^4$, such that surfaces only meet at lines and vertices, and lines only meet at vertices. Generalizing planarity, one can ask, for example, when is a 2-graph "spatial", meaning that it can be embedded in $mathbb{R}^3$.

I gather that 3-dimensional CW-complexes should correpond to such realizations of spatial 2-graphs, but find it hard to visualize the situation. Moreover, for practical porpouses I would prefer to work with the more intuitive notion of a 2-graph as a purely combinatorial object, and use their representations just as a tool for visualizing incidences, instead of working with topological spaces. So here are my questions:

• do you think the sloppy definitions above are substantially correct? In particular, is it too unfair to impose $partial^2 = 0$ rather then derive it?


• do realizations of planar graphs correspond to 3-dim cell complexes?


• is there any result such as Kuratowsky's theorem?


• can you help me find a reference where the combinatorial properties of cell-complexes (expecially 3-dim) are illustrated in detail?

Thanks

ho.history overview - What is the history of the name "Chinese remainder theorem"?

I'd be particularly interested in who first used the name in a European language and whether it was used in a non-European language such as Arabic, Persian, or an Indian language before that.

[Edit 2010/01/22: Thanks to everyone who responded. It took me a few days to check Jonas Meyer's references. (The discussion of the CRT is on pp 175-176 of Part III of Wylie's book.) As JM said, they seem to narrow the appearance of the name in a European language to 1853-1929, which is hundreds of years later than I expected, and it now wouldn't be so surprising if it first appeared in English, maybe even in Dickson's book. So,

Question: Are there any European languages in which the CRT has a name that is not a direct translation of "Chinese remainder theorem"?

One more point: Wylie says,

'In examining the productions of the Chinese one finds considerable difficulty in assigning the precise date for the origin of any mathematical process; for on almost every point, where we consult a native author, we find references to some still earlier work on the subject. The high veneration with which is has been customary for them to look upon the labours of the ancients, has made them more desirous of elucidating the works of their predecessors than of seeking fame in an untrodden path; so that some of their most important formulae have reached the state in which we now find them by an almost innumerable series of increments. One of the most remarkable of these is the Ta-yen, "Great Extension," a rule for the resolution of indeterminate problems. This rule is met with in embryo in Sun Tsze's Arithmetical Classic under the name of Wuh-puh-chi-soo, "Unknown Numerical Quantities," where after a general statement in four lines of rhyme the following question is proposed: ...

In tracing the course of this process we find it gradually becoming clearer till towards the end of the Sung dynasty, when the writings of Tsin Keu-chaou put us in full possession of the principle, and enable us to unravel the meaning of the above mysterious assemblage of numerals....'

The Song dynasty apparently ended in 1279, which gives an interval of several hundred years. So, it seems that the name Chinese Remainder Theorem is not completely unreasonable, since according to Wylie, it's not clear when the general form was discovered, or at least might not have been at the time the theorem got its name. ]

higher category theory - Joins of simplicial sets

It might be helpful to work through some simple examples. You probably know that Δⁿ ★ Δ^k = Δ^n+k+1. This has to do with the ordinal sum: one way of defining joins is as a restriction of the monoidal structure on augmented simplicial sets, which are contravarient functors from the category Δ₊ of all finite ordinals (including the empty ordinal) into sets. The category Δ₊ has a monoidal structure given by ordinary addition with ∅ as the unit, and this induces the aforementioned monoidal structure on augmented simplicial sets. The thing we call n when we are talking about simplicial sets is really the ordinal n+1, so the formula above holds because

(n+1) + (k+1) = (n+k+1)+1.

Of course, this example doesn't illustrate the asymmetry you asked about, but this one will:

∂Δⁿ ★ Δ⁰ = Λⁿ⁺¹[n+1] while Δ⁰ ★ ∂Δⁿ = Λ⁰[n+1].

To work out the details, you'll need to understand how the face maps of S★T are defined, as alluded to above. Here's my notation: (S★T)_n = S_n ∪ T_n ∪ (∪ _{j+k = n+1} S_j × T_k ).

The i-th boundary map d_i : (S★T)_n → (S★T)_n-1 is defined on S_n and T_n using the i-th boundary map on S and T. Given σ∈S_j and τ∈T_k , we have:

d_i (σ, τ) = (d_i σ,τ) if i ≤ j, j ≠ 0.
d_i (σ, τ) = (σ,d_i-j-1 τ) if i > j, k ≠ 0.

If j = 0, d₀(σ, τ) = τ ∈ T_n-1 ⊂ (S★T)_n-1. If k = 0, d_n(σ, τ) = σ ∈S_n-1 ⊂ (S★T)_n-1 .

Try this out for n = 1 or 2 first, to get a feel for things. While these sorts of computations can be quite annoying, I find they do really help me develop my intuition. Best of luck!

linear algebra - Simultaneous diagonalization

All of these are true. First note that the space of endomorphisms of $V$ is finite-dimensional, so even an infinite $S$ can just be replaced by finitely many matrices that have the same span (it's really more elegant to think about the span of $S$ as a Lie algebra, rather than $S$ itself). You actually may want to look at some discussion of abelian Lie algebras, since really your question is about the natural structure theorem for semi-simple representations of abelian Lie algebras (if you think in this language question 4) is obvious from the first 3, since any representation has a flag whose successive quotients are semi-simple).

The important point for proving 1) is that if A and B commute and are both diagonalizable, you should analyze the action of B on the eigenspaces of A. The spaces $E_chi$ above are the eigenspaces of B's action on each eigenspace of A (and if there were a third matrix, you would take the eigenspaces of C acting on the eigenspaces of B in the eigenspaces of A, etc.). 2) is clear, and for 3) just pick any basis of these iterated eigenspaces.

For 3) => 1), you have associated to each basis vector a $chi$, given by looking at how the elements of $S$ act on it. $E_chi$ is just the span of all vectors associated to the particular map $chi$.

For 4), there is a similar argument, but you have to use a flag (the $i$-th subspace being things killed by $(A-lambda I)^i$ for some scalar $lambda$) rather than a subspace decomposition. Still, matrices commuting means that B will preserve this flag, so just as one refined the eigenspaces, one can refine this flag.

at.algebraic topology - Homology of Surfaces with Holes

The classification theorem for surfaces says that the complete set of homeomorphism classes of surfaces is

{ $S_g : g geq 0$ } $ cup$ { $N_k : k geq 1$ },

where $S_g$ is a sphere with $g$ handles, and $N_k$ is a sphere with $k$ crosscaps. The first homology groups are easy to compute. They are $H_1 (S_g) = mathbb{Z}^{2g}$, and $H_1 (N_k)=mathbb{Z}^{k-1} times mathbb{Z} / 2mathbb{Z}$. My question concerns how the homology groups change once we start cutting holes in our surface.

In the orientable case, it is easy to see what happens. The first hole that we cut out does not change the homology. Every additional hole then introduces another factor of $mathbb{Z}$.

In the non-orientable case something peculiar happens. Consider the projective plane with homology group $mathbb{Z} / 2 mathbb{Z}$. If I cut out a hole, then I get the Mobius strip, which has homology group $mathbb{Z}$ (it is homotopic to a circle). In general, if I cut a hole out of $N_k$, then in the homology group I lose a factor of $mathbb{Z} / 2 mathbb{Z}$, and introduce a factor of $mathbb{Z}$. Each additional hole will then just introduce another factor of $mathbb{Z}$.

My question: In the non-orientable case what happened to the factor of $mathbb{Z} / 2 mathbb{Z}$? Is there a nice geometric explanation of why it went away? I'm slightly disturbed because I had the intuition that torsion was supposed to record non-orientability, but I guess this doesn't work for surfaces with holes.

ag.algebraic geometry - Some questions on the intersection theory on a Hilbert scheme of points of a surface.

If $Sigma$ is a smooth complex curve in a smooth projective surface $X$, then we can consider the homology class represented by $Sigma^{[n]} subset X^{[n]}$. $ $ Where, $X^{[n]}, Sigma^{[n]}$ stand for the Hilbert scheme of $n$-points on $X$ and $Sigma$, respectively. Is it possible to construct a homomorphism function $Phi_n: rm{H}_2(X) rightarrow H_w(X^{[n]})$, such that $[Sigma] mapsto [ Sigma^{[n]} ]$?

$ $ One has the following at ones disposal: we have the obvious quotient map $X^n rightarrow S^nX$ (where $S^nX$ is the symmetric product of $X$). Now, if $beta in H_2(X)$, then we can consider the image of $B := beta times cdots times beta$ in $H_{2n}(S^nX)$. If $beta $ can be represented by an algebraic curve, we can take the proper transform of $B$ under the Chow map $X^{[n]} rightarrow S^nX$. If $beta$ is not represented by such a curve, is there anything akin to proper transform that one can apply to $B$ to construct the desired homomorphism function $Phi_n$?

I am interested in studying the intersection theory between the classes $Phi_n(beta)$. Nakajima in his book "Lectures on Hilbert schemes of points on surfaces" states the following nice result. If $Sigma$ and $Sigma'$ are two smooth curves in $X$, then (page 102 of Nakajima's book):

$$sum_n z^n [Sigma^{[n]}] cdot [Sigma'^{[n]}] = (1+z)^{[Sigma] cdot [Sigma']}$$

Does anyone know if there are related results for singular curves?

As a side remark. the above formula is obvious if $Sigma$ and $Sigma'$ are two curves intersecting transversely. All it says is that of the set of $m = [Sigma]cdot [Sigma']$ points were it intersects, we choose $n$ of them (there are $binom{m}{n}$ of these guys, which is what the formula is giving). But the general proof of the formula is more intricate - one uses a representation of the Heisenberg group on the space $oplus_n H_*(X^{[n]})$ to derive it. This fancy shmancy approach is more helpful when computing things like the self intersection of $Sigma^{[n]}$ when $Sigma$ is a $(-1)$-curve in $X$. From it we get that $[Sigma^{[n]}] cdot[Sigma^{[n]}] = binom{-1}{n} = (-1)^n$

EDITED: In view of Nakajima's comment below, please replace function for homomorphism when reading the above question. Notice that, as stated in my comment below, the extension of the map $[Sigma] rightarrow [Sigma^{[n]}]$ should be a "nice" one.

EDITED (I am copying my hidden comments here since their maths don't display well)
I can explain my motivation. I am working with some moduli spaces of objects on a surface $X$ and out of them I get a homology class $V_n$ in $X^{[n]}$. In nice cases, one can show that these homology classes are $[Sigma^{[n]}]$, for some curve $Sigma subset X$. Or a sum of such classes. Using this classes $V_n$ I am trying to obtain a map $N : H_2(X) rightarrow mathbb{Z}$, defined by $N(beta) := V_n cdot Phi_n(beta)$. Such that, in the nice case when $V_n = [Sigma^{[n]}]$ and $beta = [Sigma']$, then $$N(beta) = [Sigma^{[n]}] cdot [Sigma'^{[n]}]$$
Then, my problem became what should be the definition of $Phi_n(beta)$, when $beta$ not represented by a curve. Presumably, we should be able to extend $Phi_n$ to some 2-classes that are not represented by curves since, by perturbing the complex structure, we could start seeing more curves than before. I don't know what should be $Phi_n(-2H)$. The best I could imagine is that it should satisfy the equation $$[Sigma^{[n]}] cdot Phi_n(-2H) = binom{Sigma cdot (-2H)}{n} $$ but I really don't know what it should be. Thanks a lot again!

EDIT I am now assume that the formula
$$alpha mapsto expleft( sum frac{z_i P_alpha[-i]}{(-1)^{i-1}i} right) cdot 1 $$
(the definition of the term $P_alpha[-i]$ can be found in Prof. Nakajima's book "Lectures on Hilbert schemes of points on surfaces" page 84), is well defined. By one of his results, $[Sigma] mapsto sum z^i [Sigma^{[n]}]$ (op. cit. page 99). If so, I presume this satisfy the posed question.

st.statistics - ANOVA analysis with no homogeneity of variances

I think you can still use a t-test, but you have to keep in mind that what you call "an effect" can also appear in the variance. As a consequence, I think you should add a test to your t-test to measure difference in the variances.

Simple example, consider the case when $X$ is a $mathcal{N}(0,1)$ and $Y$ is a $mathcal{N}(mu,sigma)$ and you want to say if $Y$ is the result of "an effect" or it is not.

You can look at a t-test like that:
$frac{|bar{X}-bar{Y}|}{sqrt{var(X)+var(Y)}}$. It works fine even with different variances but it will only check for differences in the mean (i.e $muneq 1$). If you want to test something on the variances also, you will have to test simultaneously (take care of the level of your test) the equality of the mean and the equality of the variances.

More General setting If an effect can appear in a more complex way, you may be obliged to concider a non parametric godness of fit testing procedure to which belong (I guess) the rank test.

Hop this help !

matrices - inverse m-matrix

Let me begin by admitting that I have no knowledge of the subject area in question: the following is my answer as a googlist, not a mathematician.

Having disclosed that, it seems that the result that you want can be found in Section 2.5 of Topics in Matrix Analysis by Horn and [C.R.] Johnson.

The extent of my grasp of the material at the moment is the following: an inverse $M$-matrix is an invertible matrix whose inverse is an $M$-matrix.

Addendum: Caveat lector: I looked more closely at the section in question and found the following passage (starting at the bottom of p. 119):

"Exercise. Show that: a) The principal minors of an inverse $M$-matrix are positive; b) every principal submatrix of an inverse $M$-matrix is an inverse $M$-matrix; and c)...Verification of these facts requires some effort."

So one sees the limits of the approach of googling and then quoting from texts: sometimes one has to put some thought into the matter! Probably someone else will enjoy reading this section of the book and working out the exercise, so I'll leave this response up, although it is certainly not an answer in and of itself.

Can skeleta simplify category theory?

The first, and perhaps most important, point is that hardly any categories that occur in nature are skeletal. The axiom of choice implies that every category is equivalent to a skeletal one, but such a skeleton is usually artifical and non-canonical. Thus, even if using skeletal categories simplified category theory, it would not mean that the subtleties were artifical, but rather that the naturally occurring subtleties could be removed by an artificial construction (the skeleton).

In fact, however, skeletons don't actually simplify much of anything in category theory. It is true, for instance, that any functor between skeletal categories which is part of an equivalence of categories is actually an isomorphism of categories. However, this isn't really useful because, as mentioned above, most interesting categories are not skeletal. So in practice, one would either still have to deal either with equivalences of categories, or be constantly replacing categories by equivalent skeletal ones, which is even more tedious (and you'd still need the notion of "equivalence" in order to know what it means to replace a category by an "equivalent" skeletal one).

In all the other examples you mention, skeletal categories don't even simplify things that much. In general, not every pseudofunctor between 2-categories is equivalent to a strict functor, and skeletality won't help you here. Even if the hom-categories of your 2-categories are skeletal, there can still be pseudofunctors that aren't equivalent to strict ones, because the data of a pseudofunctor includes coherence isomorphisms that may not be identities. Similarly for cloven and split fibrations. A similar question was raised in the query box here: important data can be encoded in coherence isomorphisms even when they are automorphisms.

The argument in CWM mentioned by Leonid is another good example of the uselessness of skeletons. Here's one final one that's bitten me in the past. You mention that universal objects are unique only up to (unique specified) isomorphism. So one might think that in a skeletal category, universal objects would be unique on the nose. This is actually false, because a universal object is not just an object, but an object together with data exhibiting its universal property, and a single object can have a given universal property in more than one way.

For instance, a product of objects A and B is an object P together with projections P→A and P→B satisfying a universal property. If Q is another object with projections Q→A and Q→B and the same property, then from the universal properties we obtain a unique specified isomorphism P≅Q. Now if the category is skeletal, then we must have P=Q, but that doesn't mean the isomorphism P≅Q is the identity. In fact, if P is a product of A and B with the projections P→A and P→B, then composing these two projections with any automorphism of P produces another product of A and B, which happens to have the same vertex object P but has different projections. So assuming that your category is skeletal doesn't actually make anything any more unique.

ag.algebraic geometry - Near Trivial Quiver Varieties

I can just tell you what the space looks like up to the action. The orbit classification of $text{hom}(V,W) oplus text{hom}(W,V)$ looks a lot like the orbit classification of $text{hom}(V,V)$, which as you know from linear algebra is given by Jordan canonical form. In fact, if you compose the two homs, the classification is clearly at least as complicated as JCF. There is just a modest amount more structure because the nilpotent part is more complicated. Although the formal dimension of the quotient is indeed negative or 0, its geometric dimension is strictly positive when $V$ and $W$ are non-trivial.

Let
$$f:V to W qquad g:W to V$$
be the two linear maps. Then $f$ has a kernel, $g circ f$ could have a larger kernel, etc. Define the stable kernel $V_0$ and the stable image $V_1$ of $g circ f$ to be the direct limits of the kernel and the image of $(g circ f)^n$. As in one proof of Jordan canonical form, $V = V_0 oplus V_1$. Similarly $W = W_0 oplus W_1$. The pair $(f,g)$ canonically splits into two pairs, $(f_0,g_0)$ and $(f_1,g_1)$. The pair $(f_0,g_0)$ is nilpotent, while the pair $(f_1,g_1)$ is invertible and establishes an isomorhism $V_1 cong W_1$.

Because of the isomorphisms between $V_1$ and $W_1$, the invariant information in the pair $(f_1,g_1)$ is the Jordan canonical form of $g_1 circ f_1$, which is the same as the JCF of the other composition. In other words, either $f_1$ or $g_1$ can be any isomorphism, and then the other one can be chosen to establish a prescribed Jordan canonical form. Any eigenvalue can appear other than 0.

The nilpotent pair $(f_0,g_0)$ is a little more interesting. It looks like an Ouroboros. An indecomposable nilpotent pair is any finite chain
$$0 to k to k to cdots to k to 0,$$
rolled up from $mathbb{Z}$-graded to $mathbb{Z}/2$-graded. (The connecting maps in the middle are all isomorphisms, not differentials.) The chains can have odd length, so that $V_1$ and $W_1$ don't have to have the same dimension. A chain of any length can also descend in two different ways to $V_1$ and $W_1$.

---

I gather from Ben's comment that this is a right answer to a wrong question. It is a good description of the representation variety of a cyclic quiver; there is nothing special about cycle length 2 in the analysis. But the $A_2$ quiver variety is something else.

big list - Which journals publish expository work?

I wonder if anyone else has noticed that the market for expository papers in mathematics is very narrow (more so than it used to be, perhaps).

Are there any journals which publish expository work, especially at the "intermediate" level? By intermediate, I mean neither (i) aimed at an audience of students, especially undergraduate students (e.g. Mathematics Magazine) nor (ii) surveys of entire fields of mathematics and/or descriptions of spectacular new results written by veteran experts in the field (e.g. the Bulletin, the Notices).

Let me give some examples from my own writing, mostly just to fix ideas. (I do not mean to complain.)

1) About six years ago I submitted an expository paper "On the discrete geometry of Chicken McNuggets" to the American Mathematical Monthly. The point of the paper was to illustrate the utility of simple reasoning about lattices in Euclidean space to give a proof of Schur's Theorem on the number of representations of an integer by a linear form in positive integers. The paper was rejected; one reviewer said something like (I paraphrase) "I have the feeling that this would be a rather routine result for someone versed in the geometry of numbers." This shows that the paper was not being viewed as expository -- i.e., a work whose goal is the presentation of a known result in a way which will make it accessible and appealing to a broader audience. I shared the news with my officemate at the time, Dr. Gil Alon, and he found the topic interesting. Together we "researchized" the paper by working a little harder and proving some (apparently) new exact formulas for representation numbers. This new version was accepted by the Journal of Integer Sequences:

http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Clark/clark80.html

This is not a sad story for me overall because I learned more about the problem ("The Diophantine Problem of Frobenius") in writing the second version with Gil. But still, something is lost: the first version was a writeup of a talk that I have given to advanced undergraduate / basic graduate audiences at several places. For a long time, this was my "general audience" talk, and it worked at getting people involved and interested: people always came up to me afterward with further questions and suggested improvements, much more so than any arithmetic geometry talk I have ever given. The main result in our JIS paper is unfortunately a little technical [not deep, not sophisticated; just technical: lots of gcd's and inverses modulo various things] to state, and although I have recommended to several students to read this paper, so far nothing has come of it.

2) A few years ago I managed to reprove a theorem of Luther Claborn (every abelian group is isomorphic to the class group of some Dedekind domain) by using elliptic curves along the lines of a suggestion by Michael Rosen (who reproved the result in the countable case). I asked around and was advised to submit the paper to L'Enseignement Mathematique. In my writeup, I made the conscious decision to write the paper in an expository way: that is, I included a lot of background material and explained connections between the various results, even things which were not directly related to the theorem in question. The paper was accepted; but the referee made it clear that s/he would have preferred a more streamlined, research oriented approach. Thus EM, despite its name ("Mathematical Education"), seems to be primarily a research journal (which likes papers taking new looks at old or easily stated problems: it's certainly a good journal and I'm proud to be published in it), and I was able to smuggle in some exposition under the cover of a new research result.

3) I have an expository paper on factorization in integral domains:

http://math.uga.edu/~pete/factorization.pdf

[Added: And a newer version: http://math.uga.edu/~pete/factorization2010.pdf.]

It is not finished and not completely polished, but it has been circulating around the internet for about a year now. Again, this completely expository paper has attracted more attention than most of my research papers. Sometimes people talk about it as though it were a preprint or an actual paper, but it isn't: I do not know of any journal that would publish a 30 page paper giving an intermediate-level exposition of the theory of factorization in integral domains. Is there such a journal?

Added: In my factorization paper, I build on similar expositions by the leading algebraists P. Samuel and P.M. Cohn. I think these two papers, published in 1968 and 1973, are both excellent examples of the sort of "intermediate exposition" I have in mind (closer to the high end of the range, but still intermediate: one of the main results Samuel discusses, Nagata's Theorem, was published in 1957 so was not exactly hot off the presses when Samuel wrote his article). Both articles were published by the American Mathematical Monthly! I don't think the Monthly would publish either of them nowadays.

Added: I have recently submitted a paper to the Monthly:

http://math.uga.edu/~pete/coveringnumbersv2.pdf

(By another coincidence, this paper is a mildly souped up answer to MO question #26. But I did the "research" on this paper in the lonely pre-MO days of 2008.)

Looking at this paper helps me to see that the line between research and exposition can be blurry. I think it is primarily an expository paper -- in that the emphasis is on the presentation of the results rather than the results themselves -- but I didn't have the guts to submit it anywhere without claiming some small research novelty: "The computation of the irredundant linear covering number appears to be new." I'll let you know what happens to it.

(Added: it was accepted by the Monthly.)

soft question - What are some examples of colorful language in serious mathematics papers?

André Weil uses some very colourful language in the introduction of his 1946 book Foundations of Algebraic Geometry. I recommend any mathematician to read it. Here are some excerpts:

"As in other kinds of war, so in this bloodless battle with an ever retreating foe which it is our good luck to be waging, it is possible for the advancing army to outrun its services of supply and incur disaster unless it waits for the quartermaster to perform his inglorious but indispensable task."

"Of course every mathematician has a right to his own language---at the risk of not being understood; and the use sometimes made of this right by our contemporaries almost suggests that the same fate is being prepared for mathematics as once befell, at Babel, another of man's great achievements."

"... however grateful we algebraic geometers should be to the modern algebraic school for lending us temporary accommodation, makeshift constructions full of rings, ideals and valuations, in which some of us feel in constant danger of getting lost, our wish and aim must be to return at the earliest possible moment to the palaces which are ours by birthright, to consolidate shaky foundations, to provide roofs where they are missing, to finish, in harmony with the portions already existing, what has been left undone."

"...it is hoped that these may be helpful to the reader, to whom the author, having acted as his pilot until this point, heartily wishes Godspeed on his sailing away from the axiomatic shore, further and further into open sea."

at.algebraic topology - Naive Z/2-spectrum structure on E smash E?

Given a spectrum $E$ there is a "standard" lift of $E wedge E$ to a $mathbb{Z}/2$-spectrum using the basic technique you describe. One way to describe it as follows.

You can construct the category of genuine $mathbb{Z}/2$-spectra (indexed on the full universe) via collections of $mathbb{Z}/2$-spaces $X_n$ with equivariant structure maps $sigma: S^V wedge X_n to X_{n+1}$, where $S^V = S^1 wedge S^1$ is the 1-point compactification of the regular representation $mathbb{R} times mathbb{R}$ with the "flip" action. Under this description, if $E$ is a spectrum made up of spaces $E_n$ and structure maps $S^1 wedge E_n to E_{n+1}$, then you can construct $E wedge E$ as a genuine $mathbb{Z}/2$-spectrum with spaces $E_n wedge E_n$ and structure maps $S^1 wedge S^1 wedge E_n wedge E_n$ that simply twist and apply the structure map on each factor. (This, e.g., is one way to pass forward the equivariant structure on $TC$).

This fully genuine spectrum has an underlying spectrum indexed on the trivial subgroup. The $mathbb{Z}/2$-fixed point object is the homotopy pullback of a diagram

$$
(E wedge E)^{hmathbb{Z}/2} to (E wedge E)^{tmathbb{Z}/2} leftarrow E
$$

where $Z^{tmathbb{Z}/2}$ is the so-called "Tate spectrum" of $Z$, which is to Tate cohomology as the homotopy fixed point spectrum is to group cohomology.

If $E = Sigma^infty W$ for a space $W$ then the map from $E$ to the Tate spectrum lifts (via the diagonal) to a map to the homotopy fixed point spectrum, and so the homotopy pullback will actually be homotopy equivalent to $E vee (E wedge E)_{hmathbb{Z}/2}$. (This homotopy orbit is the fiber of the map from homotopy fixed points to Tate fixed points.)

For finite spectra the map from $E$ to its Tate spectrum is 2-adic completion; this is one way to state the content of the Segal conjecture that Carlsson proved (at least at the prime 2). Sverre Lunoe-Nielsen extended this result to a number of other spectra like the Brown-Peterson spectra. In these cases the fixed-point object is equivalent after 2-adic completion to the homotopy fixed point object.

All the above plays out the same way for a cyclic group of prime order.

teaching - How to teach addition of negative numbers?

There are some nice physical models of integers that I learned about when I was working with elementary school teachers. One that I find particularly useful is the following:

You have two kinds of chips (say, red and black, to correspond with the standard accounting practice). The rule is that a red chip cancels the black chip. If after that, if you have k black chips, that represents the positive integer k; if you have k red chips, that represents the negative integer -k, and if you got nothing, that's 0.

The first thing that a student needs to figure out is that there are a lot of ways to represent any integer, cause you can always add a pair of red and black chips. This eventually gets the point across that adding $(k-k)$ doesn't change the value.

Addition of two integers is easy: put both piles of chips together and figure out what integer you got.

Subtraction is not much more difficult: it's a take-away operation, analogous to what people are familiar with in counting numbers.
To do $a-b$ as a take-away operation, but you may need to modify $a$ by adding more pairs of chips to have enough to take away $b$ from it.

For example: 2-5
2 is represented five black and three red chips. Take away five black chips that represent 5, and you are left with three red chips which represent -3.

The other advantage of this model is that it's easy to demonstrate that $a-b=a+(-b)$. For example, consider 2-5 as 2+(-5): two black chips, put together with five red ones, which after cancellation yield three red chips representing -3.

lo.logic - What is a reference for an explicit, logic-based, statement of duality in category theory (in ''complicated'' situations)? And what are the prerequisites for a beginner in logic?

Background

In the course of reading Mac Lane linearly (currently in Chapter VI),
I have seen again and again that duality can make life much easier. My
problem is that I have almost no background in logic, and duality is a
theorem in logic about category theory.

When I first read about duality in Chapter II of Mac Lane in the
context of the elementary theory of a single category, everything was
pretty clear even without knowing any logic. However, when I got to
the chapter on adjunctions, involving two categories and functors
between them, a bijection of hom-sets, and two natural
transformations, I got confused to the point that I wasn't even sure
how to use duality (let alone, why it is correct).

At this stage, I made a rather long pause and read the first three
chapters of Ebbinghaus, Flum, and Thomas' ''Mathematical logic'' (so, I
have read about the syntax and semantics of first-order logic). From
this, I built my own (hopefully correct) ''poor man's proof of
duality'' up to the situation of a single adjunction. This has both
clarified the validity of duality for formulas involving adjunctions,
and helped me understand how to use duality in such situations.

But a single adjunction is far from the most ''complicated'' situation
one meets. There are composition of adjunctions, pointwise limits in
functor categories, and many other situations in which I am still
not totally convinced that I understand duality (both theoretically
and practically).

For example, in one answer to a recent question
on pointwise limits in functor categories, it was stated that the
reference for limits is Mac Lane, while the reference for colimits
is Mac Lane--Moerdijk. I really wanted to comment that the
assertion on colimits is just the dual of the one on limits, but then
I realized that I am not totally sure. I would be most grateful for some solid source that
I can consult whenever I have doubts in what I get after doing the intuitive
things (reverse arrows but not functors, etc.).

Questions

1. What is a good reference for an explicit, logic-based, statement of a duality
   theorem of category theory in ''complicated situations?''


2. What are the prerequisites in logic? For example, up to which
   point of Ebbinghaus--Flum--Thomas should I read?

ag.algebraic geometry - Non-simply-connected smooth proper scheme over Z?

I think I have an argument that might work. The goal is to prove that this is impossible. There are some gaps in it.

Let $X$ be a connected smooth proper scheme over $mathbb Z$. Clearly $Gamma(X,mathcal O_X)=mathbb Z$. (If the ring had zero-divsors, it would indicate $X$ reducible, impossible, or $X$ non-reduced, thus ramified, impossible. If it were a ring of integers of a number field it would give ramification at some prime.) Since $H^1(X,mathcal O_X)$ is the tangent space of the Picard scheme, and the Picard scheme is trivial, $H^1(X,mathcal O_X)$ is trivial. (This probably requires smoothness of the Picard scheme. I'm not sure if that holds.) I need to assume that $H^2(X,mathcal O_X)$ is torsion-free. (I would think that smoothness over a scheme should imply locally free higher pushforwards, which over an affine scheme implies torsion-free cohomology, but I don't know. This is true in characteristic 0 by Deligne, but we are obviously not in characteristic 0 here.)

We have the exact sequence $0to mathcal O_X to mathcal O_X to mathcal O_X/pto 0$, with the first map multiplication by $p$. Taking cohomology and filling in what we know, we get

$ 0 to mathbb Z to mathbb Z to H^0(X, mathcal O_X/p) to 0 to 0 to H^1(X,mathcal O_X/p) to H^2(X,mathcal O_X)to H^2(X,mathcal O_X) $

which since those are also the cohomology groups of $X_P$, gives $Gamma(X_p,mathcal O_{X_p})=mathbb F_p$, $H^1(X_p,mathcal O_{X_p})=0$.

Now let $Yto X$ be a cyclic etale cover of degree $p$. Artin-Schreier on $X$ gives $H^1_{et}(X_P,mathbb Z/p)=mathbb Z/p$. Thus there is a unique connected etale degree-$p$ cover of $X_p$, so it's the one you get by tensoring over $mathbb F_p$ with $mathbb F_{p^p}$. Since $Gamma(Y_p,mathcal O_{Y_p})=mathbb F_p$, it is connected, and is not the result of tensoring anything with $mathbb F_{p^p}$. This is a contradiction.

No cyclic etale covers of degree $p$ $implies$ no cyclic etale covers $implies$ no etale covers. (since ever group has a cyclic subgroup.)

soft question - Taking lecture notes in lectures

Note taking is a bit of a religious dogma for me.

As a chemistry major,I was trained by Dr.Robert Engel something I've found to be very true as a student and has been confirmed by educational psychologists: "There's a connection between your hand and your brain." i.e. writing something out in detail forces your brain to process it.If the notes are good and informative,I find this is very true. Indeed,a good measure of how instructive lecture notes on a subject are is how well you learned from them by taking them down in detail!

That's why to be honest,I'm a little shocked by the responses here to the effect that note taking distracts them from a lecture. How can you be distracted from what the speaker in a graduate level mathematics colloquia is saying if you're forcing yourself to take notes on it?!? Yes,speakers in real time go quickly and of course,we're usually not completely alert and awake-but doesn't that force the mind to pay attention more?

I take very detailed notes along with my commentary. And later-I dissect the notes exhaustively. Or to use Paul Halmos' words;"Don't just read it,FIGHT IT!" And I do. Fiercely. I can think of no better advice to give your students.

There's also a more personal reason for taking notes in a detailed way:Each set of notes is a living record of an experience in your life. It's also a documentation of a personal style of a lecturer-each set of notes is like a personal fingerprint of the author. That relic will remain with the note taker long after the lecture ends.Your memories of the experience will be forever intertwined with those notes.

Now a lot of people here-such as Anton and Theo-are making the case for detailed TeX-ing of notes.Looking at their creations at thier websites as well as the notes made up by my other fellow graduate students,I have to admit-they're making a very compelling case for it. I just wonder about whether or not that personalized element so conducive to learning and nostalgia will be lost once everyone does this.

But once again-the results are VERY impressive. So it's certainly worth a good hard thought.

Those are my 2 cents on the issue.

ag.algebraic geometry - Frobenius Descent

Let $S$ be a scheme of positive characteristic $p$ and $X$ a smooth $S$-scheme. Let $F:Xrightarrow X^{(p)}$ denote the relative Frobenius. A result by Cartier (often called Cartier descent or Frobenius descent) then states that the category of quasi-coherent $mathcal{O}_{X^{(p)}}$-modules is equivalent to the category of quasi-coherent $mathcal{O}_X$-modules $(E,nabla)$ with integrable connection of $p$-curvature $0$ (which means $nabla(D)^p-nabla(D^p)=0$ for all $S$-derivations $D:mathcal{O}_Xrightarrow mathcal{O}_X$).
The equivalence is given by

$$ (E,nabla)longmapsto E^nabla$$

and

$$ Emapsto (F^*E,nabla^{can})$$

where $nabla^{can}$ is the canonical connection locally given by $fotimes smapsto (1otimes s)otimes df$, for

$$fotimes sin mathcal{O}_X(U)otimes E(U).$$
(tensor over the sections of the structure sheaf of $X^{(p)}$, somehow jtex can't handle that)

The proof of this theorem can be found in 5.1. in Katz' paper "Nilpotent connections and the monodromy theorem"

My question is: As $X/S$ is smooth, the relative Frobenius is faithfully flat (at least it is if $S$ is the spectrum of a perfect field), can the above theorem be interpreted as an instance of faithfully flat descent along $F$? In other words, does the connection $nabla$ give rise to a descent datum for $E$ with respect to $F$?

I know that connections are "first-order descent data", i.e. modules with connection descend along first order thickenings, but I don't see how this applies here.

fa.functional analysis - Orthonormal basis for non-separable inner-product space

Suppose X is an inner product space, with Hilbert space completion H (actually, I'm interested in the real scalar case, but I doubt there's any difference). If H is separable, then so is X, and I can find a (countable or finite) orthonormal basis of H inside X. Indeed, start with some countable subset Y of X which is dense in H. Then, by induction, we can move to a linearly independent subset of Y, and then apply Gram–Schmidt, again by induction. The point (to me, anyway) is that at any stage, we never take limits, and so we never leave X.

Now, what happens if H is not assumed separable? I've tried to use a Zorn's Lemma argument, but I keep end up wanting to take limits (or, rather, infinite sums) which gives me an orthonormal basis (in the generalised, non-countable, sense) in H, but I cannot ensure that it's in X. Am I just missing something obvious, or is there a slight technicality here...?

gn.general topology - nonhausdorff dimension

What you have defined is what general topologists call an ordinal invariant of a topological space. This is just what it sounds like: an assignment of an ordinal number to every topological space in such a way that homeomorphic spaces get assigned equal ordinals.

Knowing this terminology may help you search the literature to see whether your invariant already appears. For instance, you might look here:

---

Kannan, V.
Ordinal invariants in topology.
Mem. Amer. Math. Soc. 32 (1981), no. 245, v+164 pp.

MathReview by S.P. Franklin:

What follows is the text of an advertising blurb for this manuscript used by the AMS. Filtering out the obvious sales pitch leaves a general but accurate account of the contents: ``The concept of the order of a map is so powerful as to form a base for the unification of several ordinal invariants in topology. In this work, the author shows that the derived length of scattered spaces, sequential order of sequential spaces, etc., can all be described in terms of this notion. This view helps to extend them so as to be defined for all topological spaces without missing their most significant properties, to dualize them, to perceive them in the background of category theory and to obtain a lot of new information. In this self-contained work the author incidentally comes across some close connections among such apparently unrelated areas of topology as Čech closures, coreflective subcategories, special morphisms and the ordinal invariants mentioned above. The notion of $E$-order introduced here provides a unification of such invariants as sequential order, $k$-order, $m$-net-order and so on. This theory is not only more satisfactory than the earlier attempts of unification but also encompasses them as subcases.''

The list beginning with sequential order should also include the derived order.

---

To me it seems possible that this definition and your result -- that every ordinal number arises in this way from a topological space -- could be publishable, if written up in a succinct and attractive way.

fa.functional analysis - Is there a general notion of entropy for the states of a C*algebra?

There are several different definitions of entropy associated with operator algebras. It would be good if you knew which one you were referring to. Do you have a reference? I expect you're talking about the generalization of Kullback-Leibler relative entropy to matrix algebras.

There's a perfectly good definition of entropy for density matrices (positive self-adjoint trace 1 matrices); this is called von Neumann entropy, and although I haven't thought about it I think you should be able to extend it to a definition of entropy for self-adjoint trace 1 operators for II$_1$ factors in von Neumann algebras in the same way that you can extend Shannon entropy to continuous distributions by using differential entropy. In fact, I'd be surprised if that hasn't already been done.

gr.group theory - How big can the irreps of a finite group be (over an arbitrary field)?

EDIT: Part 4 added. EDIT2: Second proof of Part 4 added.

1. The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $leq left|Gright|$. This is actually obvious: Take any nonzero vector $vin V$; then, $kleft[Gright]v$ is a nontrivial subrepresentation of $V$ of dimension $leqdimleft(kleft[Gright]right)=left|Gright|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $dim Vleqleft|Gright|$.

2. Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $leqleft|Gright|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step:

If $dim V=left|Gright|$, then the map $kleft[Gright]to V, gmapsto gv$ must be bijective (in fact, it is surjective,
since $kleft[Gright]v=V$, and it therefore must be bijective since $dimleft(kleft[Gright]right)=left|Gright|=dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $Vcong kleft[Gright]$. But $kleft[Gright]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation).

3. Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<sqrt{left|Gright|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $left|Gright|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $dim Vleq left|Gright|-1$ can be sharp (take cyclic groups).

4. There is a way to improve 2. so that it comes a bit closer to 3.:

Theorem 1. If $V_1$, $V_2$, ..., $V_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $dim V_1+dim V_2+...+dim V_mleqdim A$.

(Of course, if $A$ is the group algebra of some finite group $G$, then $dim A=left|Gright|$, and we get 2. as a consequence.)

First proof of Theorem 1. At first, for every $iinleftlbrace 1,2,...,mrightrbrace$, the (left) representation $V_i^{ast}$ of the algebra $A^{mathrm{op}}$ (this representation is defined by $acdot f=left(vmapsto fleft(avright)right)$ for any $fin V_i^{ast}$ and $ain A$) is irreducible (since $V_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{mathrm{op}}$ (since we can choose some nonzero $uin V_i^{ast}$, and then the map $A^{mathrm{op}}to V_i^{ast}$ given by $amapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V_i^{ast}$ and therefore equal to $V_i^{ast}$ due to the irreducibility of $V_i^{ast}$). Hence, by duality, $V_i$ is isomorphic to a subrepresentation of the (left) representation $A^{mathrm{op}ast}=A^{ast}$ of $A$. Hence, from now on, let's assume that $V_i$ actually is a subrepresentation of $A^{ast}$ for every $iinleftlbrace 1,2,...,mrightrbrace$.

Now, let us prove that the vector subspaces $V_1$, $V_2$, ..., $V_m$ of $A^{ast}$ are linearly disjoint, i. e., that the sum $V_1+V_2+...+V_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V_1+V_2+...+V_{m-1}$ is already a direct sum. It remains to prove that $V_mcap left(V_1+V_2+...+V_{m-1}right)=0$. In fact, assume the contrary. Then, $V_mcap left(V_1+V_2+...+V_{m-1}right)=V_m$ (since $V_mcap left(V_1+V_2+...+V_{m-1}right)$ is a nonzero subrepresentation of $V_m$, and $V_m$ is irreducible). Thus, $V_msubseteq V_1+V_2+...+V_{m-1}$. Consequently, $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$ (because the sum $V_1+V_2+...+V_{m-1}$ is a direct sum, according to our induction assumption).

Now, according to Theorem 2.2 and Remark 2.3 of Etingof's "Introduction to representation theory", any subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$ must be a direct sum of the form $r_1V_1oplus r_2V_2oplus ...oplus r_{m-1}V_{m-1}$ for some nonnegative integers $r_1$, $r_2$, ..., $r_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$ must be one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. Since we know that $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1oplus V_2oplus ...oplus V_{m-1}$, we conclude that $V_m$ is isomorphic to one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. This contradicts the non-isomorphy of the representations $V_1$, $V_2$, ..., $V_m$. Thus, we have proven that the sum $V_1+V_2+...+V_m$ is actually a direct sum. Consequently, $dim V_1+dim V_2+...+dim V_m=dimleft(V_1+V_2+...+V_mright)leq dim A^{ast}=dim A$, and Theorem 1 is proven.

Second proof of Theorem 1. I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in Crawley-Boevey's "Lectures on representation theory and invariant theory":

Let $0=A_0subseteq A_1subseteq A_2subseteq ...subseteq A_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $iin leftlbrace 1,2,...,krightrbrace$, the representation $A_i/A_{i-1}$ of $A$ is irreducible.

Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $Iin leftlbrace 1,2,...,krightrbrace$ such that $Tcong A_I/A_{I-1}$ (as representations of $A$).

In fact, let $I$ be the smallest element $iin leftlbrace 1,2,...,krightrbrace$ satisfying $A_iTneq 0$ (such elements $i$ exist, because $A_kT=AT=Tneq 0$). Then, $A_ITneq 0$, but $A_{I-1}T=0$. Now, choose some vector $tin T$ such that $A_Itneq 0$ (such a vector $t$ exists, because $A_ITneq 0$), and consider the map $f:A_Ito T$ defined by $fleft(aright)=at$ for every $ain A_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A_{I-1}$ to $0$ (because $fleft(A_{I-1}right)=A_{I-1}tsubseteq A_{I-1}T=0$), it gives rise to a map $g:A_I/A_{I-1}to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A_I/A_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A_I/A_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $gleft(A_I/A_{I-1}right)=fleft(A_Iright)=A_Itneq 0$), so that $g$ must be an isomorphism, i. e., we have $Tcong A_I/A_{I-1}$.

So we have just proven that

(1) For every irreducible representation $T$ of $A$, there exists some $Iin leftlbrace 1,2,...,krightrbrace$ such that $Tcong A_I/A_{I-1}$ (as representations of $A$).

Denote this $I$ by $I_T$ in order to make it clear that it depends on $T$. So we have $Tcong A_{I_T}/A_{I_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V_i$ for every $iinleftlbrace 1,2,...,mrightrbrace$, we see that $V_icong A_{I_{V_i}}/A_{I_{V_i}-1}$ for every $iinleftlbrace 1,2,...,mrightrbrace$. Hence, the elements $I_{V_1}$, $I_{V_2}$, ..., $I_{V_m}$ of the set $leftlbrace 1,2,...,krightrbrace$ are pairwise distinct (because $I_{V_i}=I_{V_j}$ would yield $V_icong A_{I_{V_i}}/A_{I_{V_i}-1}=A_{I_{V_j}}/A_{I_{V_j}-1}cong V_j$, but the representations $V_1$, $V_2$, ..., $V_m$ are pairwise nonisomorphic), and thus

$sumlimits_{i=1}^{m}dimleft(A_{I_{V_i}}/A_{I_{V_i}-1}right)=sumlimits_{substack{jinleftlbrace 1,2,...,krightrbrace ; \ text{there exists }\ iinleftlbrace 1,2,...,mrightrbrace \ text{ such that }j=I_{V_i}}}dimleft(A_j/A_{j-1}right)$
$leq sumlimits_{jinleftlbrace 1,2,...,krightrbrace}dimleft(A_j/A_{j-1}right)$ (since $dimleft(A_j/A_{j-1}right)geq 0$ for every $j$, so that adding more summands cannot decrease the sum)
$=sumlimits_{j=1}^{k}dimleft(A_j/A_{j-1}right)=sumlimits_{j=1}^{k}left(dim A_j-dim A_{j-1}right)$.

Since $dimleft(A_{I_{V_i}}/A_{I_{V_i}-1}right)=dim V_i$ for each $i$ (due to $A_{I_{V_i}}/A_{I_{V_i}-1}cong V_i$) and $sumlimits_{j=1}^{k}left(dim A_j-dim A_{j-1}right)=dim A$ (in fact, the sum $sumlimits_{j=1}^{k}left(dim A_j-dim A_{j-1}right)$ is a telescopic sum and simplifies to $dim A_k-dim A_0=dim A-dim 0=dim A-0=dim A$), this inequality becomes $sumlimits_{i=1}^{m}dim V_ileqdim A$. This proves Theorem 1.