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Let $G :=G(k,n)$ be the Grassmannian of $k$-planes in an $n$-dimensional vector space. We automatically have the exact sequence for the universal (tautological) bundle $\mathcal{S}$:

$$0 \to \mathcal{S} \to \mathcal{O}_{G}^{n} \to \mathcal{Q} \to 0. $$

Then we have the following description of the tangent sheaf for $G$:

$$\mathcal{T}_G \sim \mathcal{Hom} (\mathcal{S}, \mathcal{Q}).$$

The way I have seen this isomorphism is described is by basically showing that cocycles for those bundles coincide (e.g. in "3254 and all that" by Eisenbud and Harris).

I was wondering if there is a more conceptual way to show this isomorphism other than comparing cocycles?

Ideally, I'd like to know the geometric construction of this map in the algebraic category (so without germs of curves, etc.).

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The Grassmannian represents a functor. You can compute the tangent bundle by evaluating the functor on square zero nilpotent extensions.

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