# External power of locally free sheaves

I have a question about exercise II. 5.16 (d) from R. Hartshornes Algebraic Geometry:

Now let $$(X,O_X)$$ be a ringed space and let $$0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$$ be an exact sequence of locally free sheaves $$\mathcal{F}',\mathcal{F},\mathcal{F}''$$ of constant ranks $$n',n,n''$$.

I have to show isomorphism $$\bigwedge^n \mathcal{F} \cong \bigwedge^{n'} \mathcal{F}' \otimes \bigwedge^{n''} \mathcal{F}''$$

My apprach is: take open subset $$U \subset X$$ such that $$\mathcal{F}' \vert _U,\mathcal{F} \vert _U,\mathcal{F}'' \vert _U$$ are free, then $$\mathcal{F} \vert _U \cong \mathcal{F}' \vert _U \oplus \mathcal{F}'' \vert _U$$. This is because sequences of free sheaves split. Then $$\bigwedge^r \mathcal{F} \vert _U = \bigoplus_{p=0}^r (\bigwedge^p \mathcal{F}' \vert _U \otimes \bigwedge^{r-p} \mathcal{F}'' \vert _U)$$. But if $$r > n$$, we have $$\bigwedge^r \mathcal{F} \vert _U=0$$ and same for $$\mathcal{F}'$$ and $$\mathcal{F}$$, so $$\bigwedge^n \mathcal{F} \vert _U = \bigwedge^{n'} \mathcal{F}' \vert _U \otimes \bigwedge^{n''} \mathcal{F}'' \vert _U$$ because the rest terms are all zero.

Recall that since locally we obtained $$\mathcal{F} \vert _U \cong \mathcal{F}' \vert _U \oplus \mathcal{F}'' \vert _U$$ as $$\mathcal{F}' \vert _U,\mathcal{F} \vert _U,\mathcal{F}'' \vert _U$$ are free and therefore the exact sequence $$0 \to \mathcal{F}'\vert _U \to \mathcal{F}\vert _U \to \mathcal{F}''\vert _U \to 0$$ splits. Only locally(!). Indeed, a short exact sequence of free sheaves split, a exact sequence of locally free sheaves not, see links below. And his is exactly the problem.

Now I don't know how to continue. Glueing together the pieces $$\mathcal{F} _{U_i} \cong \mathcal{F}' \vert _{U_i} \oplus \mathcal{F}'' _{U_i}$$ for open $$U_i$$ which cover $$X= \bigcup_i U_i$$ and where the sheaves are are affine to obtain $$\mathcal{F} \cong \mathcal{F}' \oplus \mathcal{F}''$$ seems to be wrong.

This would imply that every exact sequence $$0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$$ of locally free sheaves splits globally but this is obviously wrong, see here, here or here.

What I'm doing wrong here? The problem is that when I try to obtain $$\bigwedge^n \mathcal{F} \cong \bigwedge^{n'} \mathcal{F}' \otimes \bigwedge^{n''} \mathcal{F}''$$ by gluing I'm running in the wrong conclusion that every sequence of locally free sheaves splits globally since locally they always split. But this is nonsense as can be checked in examples in the links, e.g. the Euler sequence.

I would very thankful if somebody could explain where the problem lies and how the exercise can correctly be solved.

• The error here is a classic error when dealing with sheaves: one ought to define morphisms globally and then check locally. Hint: try to define a global map of sheaves using universal properties, and then you can check on a trivializing cover that it's an isomorphism. This second part will proceed very similarly to the work you've already done. – KReiser Apr 21 '20 at 22:31
• @KReiser:Hi, KReiser. Have you a hint for the choice of the global map we are looking for? I tried to use universal properties. We need tho morphisms $\bigwedge^{n'} \mathcal{F}' \subset \bigwedge^{\bullet} \mathcal{F}' \to \bigwedge^{\bullet} \mathcal{F}$ and $\bigwedge^{n''} \mathcal{F}'\subset \bigwedge^{\bullet} \mathcal{F}'' \to \bigwedge^{\bullet} \mathcal{F}$. Now we can use the universal property of tensor product. – user7391733 Apr 21 '20 at 22:58
• It's clear how we get $\bigwedge^{\bullet} \mathcal{F}' \to \bigwedge^{\bullet} \mathcal{F}$ by UP of external products. The $\bigwedge^{\bullet} \mathcal{F}'' \to \bigwedge^{\bullet} \mathcal{F}$ is the problem. Do you have a hint? – user7391733 Apr 21 '20 at 22:58
• @KReiser: Hi, I have to confess that I thought all day how to construct a global morphism $\bigwedge^{\bullet} \mathcal{F}'' \to \bigwedge^{\bullet} \mathcal{F}$ but now I have to surrender. Can you tell me how you get this global morphism $\bigwedge^{n'} \mathcal{F}' \otimes \bigwedge^{n''} \mathcal{F}'' \to \bigwedge^n \mathcal{F}$. I have up to what I tried before not the not the slightest idea. I'm too foolish. – user7391733 Apr 23 '20 at 1:24
• I'm sorry, I gave you a bad hint. I'm writing an answer now which should hopefully resolve this. – KReiser Apr 23 '20 at 4:21

I gave you a bad hint in the comments because I had not thought about this in a while. Let me try to make up with it by giving you a (hopefully enlightning) solution. The key issue in your attempted solution is that you're not really engaging with the method suggested by Hartshorne: we can define a filtration of $$\bigwedge^r\mathcal{F}$$ which has successive subquotients $$\bigwedge^p \mathcal{F}'\otimes\bigwedge^{r-p}\mathcal{F}''$$. We define this filtration locally, and then check that it's "independent enough" to patch together globally.

We're going to use the same strategy as part (c): we're going to define a filtration of $$\bigwedge^r \mathcal{F}$$,

$$\bigwedge^r\mathcal{F} = F^0\supseteq F^1\supseteq\cdots\supseteq F^r \supseteq F^{r+1}=0$$

which has successive quotients $$F^p/F^{p+1}=\bigwedge^p\mathcal{F}' \otimes \bigwedge^{r-p} \mathcal{F}''$$. Considering $$r=\operatorname{rank} \mathcal{F}$$, this will imply the result, as the filtration will reduce to $$\bigwedge^r \mathcal{F} \cong \bigwedge^{rk \mathcal{F}'}\mathcal{F}'\otimes \bigwedge^{rk \mathcal{F}''}\mathcal{F}''$$.

The way we construct this filtration is that we set it up in a basis-independent way on open subsets $$U\subset X$$ where each of $$\mathcal{F}',\mathcal{F},\mathcal{F}''$$ are free, and then check that this means it glues in to a filtration of the global sheaves. Fix an arbitrary such $$U$$, and choose any splitting $$\mathcal{F}|_U\cong \mathcal{F}'|_U\oplus \mathcal{F}|_U''$$. This gives us $$\bigwedge^r\mathcal{F}|_U \cong \bigoplus_{i=0}^r \left(\bigwedge^i \mathcal{F}'|_U\right)\otimes \left(\bigwedge^{r-i} \mathcal{F}''|_U\right).$$

Now we construct the filtration by induction - the idea of our strategy is to set $$F^i$$ to be those wedge products which have at least $$i$$ entries coming from a basis of $$\mathcal{F}'|_U$$. To do this, set $$F^{r+1}=0$$, and assume we've picked $$F^{j+1},\cdots,F^{r+1}$$ satisfying the requested properties. To construct $$F^j$$, consider the image of

$$\varphi: \bigwedge^j\mathcal{F}'|_U \otimes \bigwedge^{r-j} \mathcal{F}''|_U \to \left(\bigwedge^r \mathcal{F}|_U\right)/F^{j+1}.$$

I claim the preimage of this under the projection $$\pi:\bigwedge^r\mathcal{F}|_U\to \left(\bigwedge^r\mathcal{F}|_U\right)/F^{j+1}$$ is independent of the chosen splitting: pick a basis $$x_1,\cdots,x_s$$ for $$\mathcal{F}'|_U$$ and a basis $$y_1,\cdots,y_t$$ for $$\mathcal{F}''|_U$$, where we identify each with their image inside $$\mathcal{F}|_U$$. The image of $$\varphi$$ is the span of the collection $$x_{p_1}\wedge \cdots\wedge x_{p_j}\wedge y_{q_1}\wedge\cdots\wedge y_{q_{r-j}}$$ as the $$p$$ range over $$1,\cdots,s$$ and the $$q$$ range over $$1,\cdots,t$$. Then for any other basis $$y_1+c_1,\cdots,y_t+c_t$$ with $$c_i\in \mathcal{F}'|_U$$, the image is the span of the collection $$x_{p_1}\wedge \cdots\wedge x_{p_j}\wedge (y_{q_1}+c_{q_1})\wedge\cdots\wedge (y_{q_{r-j}}+c_{q_{r-j}}).$$ But expanding the wedge product in this sum, we see that it differs from a vector in our original collection by wedge products which have at least $$j+1$$ entries which are selected from the $$x$$s, and thus this difference is in $$F^{j+1}$$, so our claim about basis-independence is proven, and we can set $$F^j$$ to be this preimage.

This means our filtration $$F^j$$ is basis-independent, and in particular, it's compatible with restriction maps between open subsets where $$\mathcal{F},\mathcal{F}',\mathcal{F}''$$ are all free since any basis of the sheaves on the larger set is still a basis on the smaller set. Since any scheme is covered by such opens, this means we can patch our filtration together to an honest filtration of sheaves with the specified quotients globally, and we're done.

• In the local case, can't we just define $\mathcal{F}^{j}=\oplus_{k=0}^{r-j}\wedge^{r-k}\mathcal{F'}\otimes\wedge^{k}\mathcal{F''}$? – Joo Jul 12 at 9:09
• If we choose splits of the exact sequence on open sets $U$ and $V$ to construct isomorphism $f_{U},f_{V}:\mathcal{F}'\oplus\mathcal{F}''\rightarrow \mathcal F$, then on $U\cap V$ we have $f_{V}^{-1}f_{U}=id$ and the local constructions of $F^{j}$ as I suggested them are compatible because they are $\textit{identical}$. – Joo Jul 12 at 9:34