Stalks of exterior power

Question

Assume we have a ringed space $(X,\mathcal{O}_X)$ and an $\mathcal{O}_X$-module $\mathscr{F}$. Then I want to see that for all $x\in X$ we have an isomorphism $$(\bigwedge_{\mathcal{O}_X}^r\mathscr{F})_x\stackrel{\simeq}{\longrightarrow}\bigwedge^r_{\mathcal{O}_{X,x}}\mathscr{F}_x$$ of $\mathcal{O}_{X,x}$-modules, and I read in Goertz/Wedhorn that this follows from the fact that exterior-powers commute with direct limits for abstract modules over a ring. With this in mind, I started the computation with \begin{align*} (\bigwedge_{\mathcal{O}_X}^r\mathscr{F})_x&\cong (\bigwedge^r_{\mathcal{O}_X}\mathscr{F})^{\operatorname{pre}}_x\\ &\cong \varinjlim_{x\in U}\bigwedge^r_{\Gamma(U,\mathcal{O}_X)}\Gamma(U,\mathscr{F})\\ &\cong \bigwedge^r_{\Gamma(U,\mathcal{O}_X)}\varinjlim_{x\in U}\Gamma(U,\mathscr{F}), \end{align*} but then I got stuck because although I could go over to $\mathscr{F}_x$ in the argument of the latter exterior power, I wouldn't get rid of the $\Gamma(U,\mathcal{O}_X)$ in the index, and so I am a bit confused how to finish this off so that we get an isomorphism with respect to the $\mathcal{O}_{X,x}$-module structure.

Answer 1

The result you cite from Gortz and Wedhorn is that exterior powers commute with filtered inductive limits of modules over a fixed base ring - this requires more work to apply to your situation because your base rings are changing. Though this can be done, I prefer a different approach: let us prove that exterior powers commute with pullback for morphisms of ringed spaces, which will solve the problem after being applied to the morphism $(\{x\},\mathcal{O}_{X,x})\to (X,\mathcal{O}_X)$.

Let $f:Y\to X$ be a morphism of ringed spaces, let $\mathcal{F}$ be an $\mathcal{O}_X$-module, let $T^n(\mathcal{F})=\mathcal{F}\otimes_{\mathcal{O}_X}\cdots\otimes_{\mathcal{O}_X}\mathcal{F}$ be the $n$-fold tensor product, and let $\mathcal{K}_n(\mathcal{F})$ be the subsheaf of $T^n(\mathcal{F})$ generated by tensors with two entries equal. Then we have an exact sequence $$0\to \mathcal{K}\to T^n(\mathcal{F})\to \bigwedge^n\mathcal{F}\to 0,$$ and we can pull it back along $f$ to get a right-exact sequence $$ f^*\mathcal{K}\to f^*T^n(\mathcal{F})\to f^*\bigwedge^n\mathcal{F}\to 0.$$ As tensor products commute with arbitrary pullbacks of ringed spaces, we have that $f^*T^n(\mathcal{F})=T^n(f^*\mathcal{F})$. If we can show that $f^*\mathcal{K}$ surjects on to the subsheaf of $T^n(f^*\mathcal{F})$ generated by tensors with two entries equal, we win: this gives that $f^*\bigwedge^n\mathcal{F}=\bigwedge^nf^*\mathcal{F}$ because they're the same quotient.

But this is rather straightforward: by expanding a local section of $T^n(f^*\mathcal{F})$ with two coordinates equal in to a sum of pure tensors, it suffices to treat the case of $t=x\otimes x\otimes \cdots$ where $x=\sum a_i\otimes g_i$ for $a_i$ a local section of $f^{-1}\mathcal{F}$ and $g_i$ a local section of $\mathcal{O}_Y$ and all the entries in $\cdots$ are local sections of $f^{-1}\mathcal{F}$. Then $t= \sum_{i,j} (a_i\otimes g_i)\otimes (a_j\otimes g_j)\otimes \cdots$, and we see that $g_i^2(a_i\otimes a_i\otimes \cdots)$ and $g_ig_j(a_i\otimes a_j\otimes \cdots+a_j\otimes a_i\otimes \cdots)$ are all in the image of $f^*\mathcal{K}\to T^n(f^*\mathcal{F})$. $\blacksquare$

Answer 2

Question: "I am a bit confused how to finish this off so that we get an isomorphism with respect to the OX,x-module structure."

Answer: If $(X, \mathcal{O}_X)$ is a scheme and $\mathcal{E}$ a finite rank locally trivial sheaf on $\mathcal{E}$, let $f:Y \rightarrow X$ be a map of schemes. Tensor operations such as exterior products and symmetric products commute with pullback:

$$f^*Sym^d(\mathcal{E}) \cong Sym^d(f^*\mathcal{E})\text{ and}f^*(\wedge^r \mathcal{E}) \cong \wedge^r(f^*(\mathcal{E})).$$

Let $x\in U:=Spec(A) \subseteq X$ be an open affine set containing $x$ with $\mathcal{E}(U):=E$. If $j: Spec(\mathcal{O}_{X,x}) \rightarrow U$ and $i:U \rightarrow X$ are the canonical map, it follows

$$i^*(\wedge^r \mathcal{E})\cong \wedge^r (\mathcal{E}_U) $$

and

$$(\wedge^r \mathcal{E})_x:=( i \circ j)^*(\wedge^r(\mathcal{E}))=j^*(\wedge^r \mathcal{E}_U)=\wedge^r(j^*(\mathcal{E}_U))=\wedge^r(E_{\mathfrak{p}}) \cong \wedge^r(\mathcal{E}_x)$$

where $\mathfrak{p} \subseteq A$ is the prime ideal corresponding to $x$.

You must convince yourself that

$$B\otimes_A Sym^d_A(E) \cong Sym_B^d(B\otimes_A E)\text{ and }B\otimes_A \wedge^r_A E \cong \wedge_B^r(B\otimes_A E).$$

Hence if $B:=A_{\mathfrak{p}}$ and $\phi: A \rightarrow A_{\mathfrak{p}}$ is the canonical map you get

$$Sym^d_A(E)_{\mathfrak{p}} \cong A_{\mathfrak{p}}\otimes_A Sym^d_A(E) \cong Sym^d_{A_{\mathfrak{p}}}(A_{\mathfrak{p}}\otimes_A E) \cong$$

$$Sym^d_{A_{\mathfrak{p}}}(E_{\mathfrak{p}}).$$

At the level of sheaves you get with $U:=Spec(A)\subseteq X, \mathcal{E}(U):=\tilde{E}(U)$

$$Sym^d_{\mathcal{O}_X}(\mathcal{E})_x \cong (Sym^d_{\mathcal{O}_X}(\mathcal{E})_U)_x \cong Sym^d_A(E)_{\mathfrak{p}} \cong $$

$$ Sym^d_{A_{\mathfrak{p}}}(E_{\mathfrak{p}})\cong Sym^d_{\mathcal{O}_{X,x}}(\mathcal{E}_x)$$

where $x\in X$ is the point corresponding to $\mathfrak{p}$

A similar argument proves the case for the exterior product.

In general if $(X, \mathcal{O})$ is a locally ringed topological space and $E$ is a left $\mathcal{O}$-module, define for any open set $U \subseteq X$

$$F(E)(U):= \wedge^r_{\mathcal{O}(U)}E(U).$$

it follows $F(E)$ is a presheaf on $X$ with the same stalks as $\wedge^r E$. Let $I:=\{U\subseteq X\}_{x\in U}$ be the open sets of $X$ containing $x$. You want an isomorphism

$$lim_{U \in I}\wedge^r_{\mathcal{O}(U)} E(U) \cong \wedge^r_{lim_{U\in I}\mathcal{O}(U)}(lim_{U \in I}E(U)):=\wedge^r_{\mathcal{O}_x}E_x.$$

if the two objects

$$lim_{U \in I}\wedge^r_{\mathcal{O}(U)} E(U)\text{ and }\wedge^r_{\mathcal{O}_x}E_x$$

are to be isomorphic, you could prove that the right one satisfies the universal property of the left one. A direct limit is uniquely determined up to isomorphism by a universal property. If the two groups are isomorphic, there should be "canonical maps" between them.

Note: You must prove that you can take the direct limit on the left hand side. You must prove the system $\{\wedge^r_{\mathcal{O}(U)} E(U)\}_{U\in I}$ is a direct system of abelian groups.

There are canonical maps

$$\rho_U: \mathcal{O}(U) \rightarrow \mathcal{O}_x $$

defined by

$$\rho_U(s):=(s,U)$$

and

$$\eta_U: E(U) \rightarrow E_x$$

with

$$\eta_U(e):=(e,U)\in E_x.$$

It seems this gives for any $U$ a "canonical" map

$$\phi_U: \wedge^r_{\mathcal{O}(U)}E(U) \rightarrow \wedge^r_{\mathcal{O}_x}E_x$$

defined by

$$\phi_U(e_1\wedge \cdots \wedge e_r):=(e_1,U) \wedge\cdots \wedge (e_r,U) \in \wedge^r_{\mathcal{O}_x}E_x.$$

It could be this map "passes to the limit":

If this is the case you get a map

$$\phi_x:=lim_{U\in I}\phi_U: lim_{U\in I}\wedge^r_{\mathcal{O}(U)}E(U) \rightarrow \wedge^r_{\mathcal{O}_x}E_x.$$

and we may suspect that $\phi_x$ is an isomorphism.

Hint: Maybe there is a map in the other direction defined as follows: For any open set $U\in I$ there is a map

$$i_U: \wedge^r E(U) \rightarrow lim_{U\in I}\wedge^r E(U).$$

Given any element $z:=e_1\otimes \cdots \otimes e_r \in E_x^{\otimes r}$, choose representatives $e_i^* \in E(U)$ mapping to $e_i$. There is a map

$$\psi_U^*:E_x^{\otimes r} \rightarrow lim_{U \in I}\wedge^r E(U)$$

defined by

$$\psi_U^*(z):=i_U(e_1^*\wedge \cdots \wedge e_r^*) \in lim_{U\in I} \wedge^r E(U).$$

The map $\psi_U$ factors through the exterior product: You get a well defined map

$$ \psi_x: \wedge^r(E_x) \rightarrow lim_{U\in I}\wedge^rE(U).$$

If $\psi_x \circ \phi_x = \phi_x \circ \psi_x=Id$ the claim follows. You must check if the above constructions are well defined and give an isomorphism.

Note: When trying to prove two constructions in homological algebra are isomorphic you may in many cases construct two explixit maps - one in each direction. Here is a "list of explicit isomorphisms in homological algebra":

The "magic diagram" is cartesian

Prove $\phi$ to be isomorphism (an exercise in commutative algebra)

Form of basic open set of affine scheme: The intersection of two basic open sets.