Why does on a scheme $X$ of characteristic zero for a vector bundle $F$ (of finite rank) the operation $Sym^n(.)$ commute with taking the dual bundle $(.)^v$, i.e.
$Sym^n(F^v) \simeq Sym^n(F)^v$ canonically ?
How does the iso work explicitly?
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1 Answer
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There is always a map $Sym^{n}(F^{\vee}) \rightarrow Sym^{n}(F)^{\vee}$, regardless of the characteristic, and this map is an isomorphism in characteristic $0$. The map is induced by the natural pairing $Sym^{n}(F^{\vee}) \times Sym^{n}(F) \rightarrow \mathcal{O}_{X}$, given locally by $(\phi_{1}\cdots \phi_{n}, f_{1}\cdots f_{n}) \mapsto \sum_{\sigma \in S_{n}} \phi_{\sigma(1)}(f_{1}) \cdots \phi_{\sigma(n)}(f_{n})$.
If you write out the induced map in local coordinates, you'll see that some of the coefficients will be divisible by various primes, sometimes causing problems in positive characteristic. In characteristic zero, however, you can easily check that this map gives an isomorphism over a trivializing cover for $F$ and hence over all of $X$.
There is a similar pairing for alternating powers, by putting in appropriate signs. This map is always an isomorphism, regardless of characteristic, because no pesky coefficients appear.
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$\begingroup$ @Chris: thanks for this: first I think you mean "n" instead of "d" and in the first Φ there should be "1" instead of "i". I knew that this map would do, but I don't see where for example the problem appears. A second point which isn't clear to me is whether one has to multiply the right hand side by $1/n!$ or not... It would be great if you could explain this shortly. $\endgroup$– VeenCommented Jan 16, 2012 at 15:42
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1$\begingroup$ Fixed the indices. Thanks. To see why there is a problem in characteristic $p$, consider $n=2$ and ${\rm dim} V=2$ (so we are just working over a field). Choose a basis $v,w$ for $V$ and a dual basis $x,y$ for $V^{*}$. Then for example under the map $Sym^{2}(V^{*}) \rightarrow Sym^{2}(V)^{*}$, $x^2$ maps to the linear form which sends $v^2$ to $2$, $vw$ to $0$, and $w^2$ to $0$. Likewise for $y^2$. In characteristic $2$, this is clearly a problem. If you *don't$ divide by $1/n!$, then you can define the map in positive characteristic. In characteristic zero, this doesn't cause a problem. $\endgroup$ Commented Jan 16, 2012 at 15:59
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1$\begingroup$ About characteristic zero. I should add that you can use Nakayama's lemma to reduce the surjectivity of the map to surjectivity on the fibres of your bundle, and now that you are over a field, you don't need to invert denominators by hand to get surjectivity. If someone sees that we need any extra assumptions on our scheme, please speak up. $\endgroup$ Commented Jan 16, 2012 at 16:07
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$\begingroup$ So if I understand Chris' right, then the argument for the above map to be an iso in char 0 goes as follows: to show that it is an iso it suffices to show that it is surjective (as both bundles have same rank and any epi between them must be iso), and therefore one can check it on stalks, which by Nakayama reduces to show it on vector bundle fibers, and there we simply have the usual map on vector space level of char 0, from which we know it is an iso. The only thing we needed was that the residue fields are char 0, so no extra assumptions are necessary, I think... $\endgroup$– VeenCommented Jan 16, 2012 at 16:47
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$\begingroup$ By "characteristic $0$" you probably mean that $X$ defined over $\mathbb{Q}$. $\endgroup$ Commented Jan 16, 2012 at 16:54