vector bundles - On finite endomorphisms of $mathbf{P}^r$

The claimed computation is still wrong. Let $m equiv r mod n$, with $0 leq r < n$. Then the right answer is that
$$pi_* mathcal{O}(m) = mathcal{O}( lfloor (m+1)/n rfloor-1)^{oplus(n-r-1)} oplus mathcal{O}( lceil (m+1)/n rceil-1)^{oplus(r+1)}.$$

Let $S$ be the source $mathbb{P}^1$ and $T$ the target. As a general piece of advice, you should never identify two spaces when you don't have to. Here are three ways you could get this answer:

By Grothendieck-Riemmann-Roch

Let $L_S$ be the line bundle $mathcal{O}(1)$ on $S$ and $L_T$ the line bundle $mathcal{O}(1)$ on $T$. Let $H_S$ and $H_T$ be the hyperplane classes in $H^*(S)$ and $H^*(T)$. The chern character of $L_S^{otimes m}$ is $(1+H_S)^m = 1+m H_S$. The Todd classes of $S$ and $T$ are $1+H_S$ and $1+H_T$. So $pi_* L_S^{otimes m}$ is something with chern character
$$(1+H_T)^{-1} pi_*left( (1+m H_S) (1+H_S) right) = (1+H_T)^{-1} pi_*left( 1+(m+1) H_S right).$$
Note that $pi_* 1 = n$ and $pi_* H_S = H_T$. So we get
$$ch(pi_* mathcal{O}(m)) = (1+H_T)^{-1} (n+(m+1) H_T)=n + (m-n+1) H_T.$$
(We know that $R^1 pi_*$ vanishes because the map is finite.)

From the leading term, we see that $pi_* mathcal{O}(m)$ has rank $n$. It is not completely obvious that is torsion free. If we assume it is, then it must be of the form $bigoplus mathcal{O}(a_i)$ for some $a_1 + cdots + a_n$. We see from the above computation that $sum a_i = m-n+1$.

That's as far as we can get from GRR. Grothendieck-Riemann-Roch can only do the computation in $K$-theory, so we can't distinguish $mathcal{O}(-1) oplus mathcal{O}(1)$ from $mathcal{O}(0) oplus mathcal{O}(0)$.

Directly in K-theory

The point of Grothendieck-Riemann-Roch is that it gives a commuting diagram
$$begin{matrix} K^0(S) & longrightarrow & H^*(S) \
downarrow & & downarrow \
K^0(T) & longrightarrow & H^*(T). end{matrix}$$

I usually find that it is just as easy to do my computations directly on the left hand side of the diagram. Let $p_S$ and $p_T$ be the class of the structure sheaf of a point on $S$ and $T$. We have the short exact sequence $0 to mathcal{O}(-1) to mathcal{O}(0) to mathcal{O}_{mathrm{pt}} to 0$, so $p_S = 1-L_S^{-1}$ and $L_S = 1+p_S$. (Since $p_S^2=0$.)

Clearly, $pi_* p_S = p_T$. So
$$pi_* mathcal{O}(m) = pi_* (1+p_S)^m = pi_* (1+m p_S) = pi_* 1 + m p_T.$$

We can see that $pi_* 1$ has rank $n$; say $pi_* 1 = n+a p_T$.
Let $chi$ be pushforward to the $K$-theory of a point, better known as holomorphic Euler characteristic.
Since pushforward is functorial, the sequence of maps $S to T to mathrm{pt}$ shows that $$chi(pi_* 1) = chi(1) = 1$$
so $n+a=1$ and $a = -(n-1)$. We see that
$$pi_* mathcal{O}(m) = n + (m-n+1) p_T.$$

By direct computation

It is easy enough to do this example, and any toric example like it, directly from the definition of pushforward. As a bonus, this will tell us exactly which vector bundle we get, not just the $K$-class.

Let $S_1 cup S_2$ be the open cover of $S$ where $S_1 = mathrm{Spec} k[s]$ and $S_2 = mathrm{Spec} k[s^{-1}]$. Similarly, define $T_1$, $T_2$, $k[t]$ and $k[t^{-1}]$.

Let $e$ be a generator for the free $k[s]$ module $mathcal{O}(m)(S_1)$. Then $s^m e$ is a generator of $mathcal{O}(m)(S_2)$. By definition, $left( pi_* mathcal{O}(m) right) (T_1)$ is $mathcal{O}(m)(S_1)$ considered as a $k[t]$-module. As such, it has basis
$$ e, s e, s^2 e, ldots s^{n-1} e. quad (*)$$
Similarly, $left( pi_* mathcal{O}(m) right) (T_2)$ has basis
$$ s^m e, s^{m-1} e, ldots, s^{m-n+1} e. quad (**)$$

Reorder the lists $(*)$ and $(**)$ so that corresponding elements have exponents of $s$ which are congruent modulo $n$. To keep notation simple, I'll do the case of $m=0$. So we pair off:

• $e$ and $e$


• $s e$ and $s^{-n+1} e = t^{-1} (s e)$


• $s^2 e$ and $s^{-n+2} e = t^{-2} (s e)$
  and so forth.

There is one time that we pair $v$ with itself and $(n-1)$ times that we pair $v$ with $t^{-1} v$. So the transition matrix between our bases is diagonal with entries $(1,t^{-1}, t^{-1}, ldots, t^{-1})$ and the vector bundle is $mathcal{O}(0) oplus mathcal{O}(-1)^{n-1}$.

For general $m$, if I didn't make any errors, we get the formula at the beginning of the post.