This question is basically on applying the Grothendieck-Riemann-Roch theorem to finding a formula for the push-forward of a line bundle on $\mathbf{P}^r$ under a certain morphism. Since I have a lot of questions, let me begin with an example.

Let $\pi:\mathbf{P}^1 \longrightarrow \mathbf{P}^1$ be the morphism defined as $[x_0:x_1] \mapsto [x_0^n:x_1^n]$. Here $\mathbf{P}^1$ denotes the projective line over $\mathbf{C}$ and $n\geq 1$ is an integer. Note that $\pi$ is finite. For, it is locally given by the map $x\mapsto x^n$.

Let $X=\mathbf{P}^1$.

One can show that $\pi_\ast (\mathcal{O}(m))$ is given (up to isomorphism) by $$ \mathcal{O}(n(m+1)-1)\oplus \ldots \oplus \mathcal{O}(n(m+1)-1).$$ Here the sum is taken $n$ times. For example, $\pi_\ast \mathcal{O}_X \cong \mathcal{O}(n-1)^{\oplus n}$.

This is wrong. See David Speyer's response for the correct expression.

Now for the questions.

Q1. I would like to look at the morphism $\mathbf{P}^r\longrightarrow \mathbf{P}^r$ given by $[x_0:\ldots:x_r]\mapsto [x_0^n:\ldots:x_0^r]$. This is a finite morphism and therefore it should be possible to apply GRR in finding an expression for $\pi_\ast \mathcal{O}(m)$, right? Now, this is probably still very easy so I was wondering if there were any other results in this direction.

Q2. How would one do this without the Grothendieck-Riemann-Roch theorem? By "this", I also mean the above example for $\mathbf{P}^1$.

Q3. What is the geometric interpretation of this?

Final note. Since every vector bundle on $\mathbf{P}^1$ has a unique decomposition into twisted sheaves, we get a nice expression for $\pi_\ast \mathcal{E}$ where $\mathcal{E}$ is a vector bundle on $\mathbf{P}^1$.

  • 1
    $\begingroup$ Are you sure that the pushforward under, say, the squaring map of $\mathcal{O}$ on $\mathbb{P}^1$ is \mathcal{O}(1)$? It should be rank 2. $\endgroup$ – Charles Siegel Mar 22 '10 at 17:42
  • $\begingroup$ $\pi_* \mathcal{O}$ should be a vector bundle of rank $n$, not a line bundle, and it is quite possible to compute it from the definition of push-forward. I can write a good deal more about this, but I worry that this might be homework. How do you come to this problem? $\endgroup$ – David E Speyer Mar 22 '10 at 17:50
  • $\begingroup$ @David: I am currently writing my thesis on the Grothendieck-Riemann-Roch theorem. In order for me to get as flexible as possible with the theorem I decided to study some basic applications of it. For example, in a previous post I asked some questions on "families of Enriques surfaces". So this definitely isn't homework. In fact, it's just to strengthen my understanding of the theorem. Nobody told me to do this. The main question is actually if there is some more general way of looking at this. $\endgroup$ – Ariyan Javanpeykar Mar 22 '10 at 19:02
  • $\begingroup$ @Charles: I made a mistake. Thank you for pointing it out. I will edit it now. $\endgroup$ – Ariyan Javanpeykar Mar 22 '10 at 19:30

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.


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