# Interpretation of abstract Isomorphisms of Hom-sheaves

Let $X$ be a smooth $k-$ variety and $F$ a coherent sheaf on it. One knows that the sheaf of differentials $\Omega^1_X$ of $X$ is locally free.

Hence one has by standard isomorphisms

$Hom(F,F\otimes \Omega^1_X) \simeq H^0(X, \underline{Hom}(F,F\otimes \Omega^1_X)) \simeq H^0(X,\underline{Hom}(F,F)\otimes \Omega^1_X)\simeq End(F) \otimes \Omega^1_X(X)$

where in the Hom's I always mean $\mathcal O_X$-linear.

Now I am a bit confused: How do I practically get this Iso?

I mean: if I have an element $\phi \otimes \omega$ in the last vector space, how do I get my sheaf map $F \rightarrow F\otimes \Omega^1_X$ ? And also the converse isn't clear to me.

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Your isomorphism is true locally. Can you do it for modules over a regular ring, instead of sheaves over a smooth variety? In fact, if you do it for vector spaces over a field (!) then exactly the same idea will work in your case. – Mariano Suárez-Alvarez Nov 1 '11 at 11:09
The last isomorphism is wrong. Global Sections and tensor products don't commute. – Martin Brandenburg Nov 1 '11 at 11:26
The first isomorphism is trivial, so let me explain the second: for sheaves of modules $F,G,H$ on a ringed space there is a canonical morphism $\underline{Hom}(F,G) \otimes H \to \underline{Hom}(F,G \otimes H)$; construct it using the universal properties (or locally as Mariano suggests, but then you have to argue that it is glues). Observe that it is an isomorphism when $H$ is free of rank $1$, then when $H$ is free of an arbitrary finite rank, and finally if $H$ is locally free of finite rank. – Martin Brandenburg Nov 1 '11 at 11:30
@Martin: gluability as an immediate consequence of naturality. – Mariano Suárez-Alvarez Nov 1 '11 at 11:39
My suggestion is more than anything suggested by the fact that if one is to understand that isomorphism (indepedentily of sheaves, ring spaces, universal properties and what not) ne simply must be aware of how the exact same thing is done in the contexts of rings or even vector spaces. This is one of the many instances where ringed spaces and all the rest of the paraphernalia is simply complication totally irrelevant to the point. – Mariano Suárez-Alvarez Nov 1 '11 at 11:41

## 1 Answer

The natural morphism $End(F) \otimes H^0(X,\Omega^1_X) \to Hom(F,F\otimes\Omega^1_X)$ is indeed pretty "natural": just map $\phi \otimes \omega$ to $(s \mapsto \phi(s) \otimes \omega)$.

The inverse isomorphism is not as easy to write down because it uses dualizability of $\Omega^1_X$ (the fact that it is locally free of finite rank). Take a local basis $(dx_1,\ldots,dx_n)$ of $\Omega^1_X$, then any $\psi : F \to F\otimes \Omega^1_X$ can be written $\psi(s) = \sum \psi_i(s) \otimes dx_i$ where $\psi_i \in End(F)$. You should then just check that the $\sum_i \psi_i \otimes dx_i$ glue nicely as a global section of $End(F) \otimes \Omega^1_X$.

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Since locally the second map is the inverse of the first one, which is defined globally, the former must glue correctly. – Mariano Suárez-Alvarez Nov 2 '11 at 13:17