I have a second question today.
In Harris' "Algebraic Geometry: A First course" he constructs (on page 200) an isomorphism between the tangent space of the Grassmannians and some homomorphisms:
He begins with the cover of the Grassmannians defined in Section 6, i.e. for a (n-k)-plane $\Gamma \subset K^{n+1}$ he defines $U_{\Gamma}:= \{\Lambda'\in\mathbb G(k,n)|\ \Lambda \cap \Gamma=(0)\}$.
Then he constructs an isomorphism to $\mathbb A^{(n-k)(k+1)}$ as follows:
"Fixing a subspace $\Lambda\in U_{\Gamma}$ a subspace $\Lambda' \in U_{\Gamma}$ is the graph of a homomorphism $\phi:\Lambda\rightarrow\Gamma$, so that $U_{\Gamma}=$Hom$(\Lambda, \Gamma)$.
He then deduces $T_{\Lambda}(\mathbb G)=$Hom$(\Lambda, \Gamma)$.
This last step is not trivial for me so I tried to cut it up in several substeps, but I seem to be stuck:
We know by one definition of the projective tangent space at some point $p$ of some projective variety $X\subset \mathbb P^n$ that it is the projective closure of the affine tangent space of $X\cap U$, where $p\in U$ is isomorphic to $\mathbb A^n$.
Now for $\Lambda \in \mathbb G$ there exists such an open subset $\Lambda \in U\subset \mathbb P^{\binom{n+1}{k+1}}$ such that $\mathbb G\cap U=U_{\Gamma}$ for some $\Gamma$. Then by the fact above, we know
$T_{\Lambda}(\mathbb G)=\bar{T_{\Lambda}(U_{\Gamma})}=\bar{U_{\Gamma}}$, as $U_{\Gamma}$ is affine (the bars are supposed to denote the projective closure). How do I continue?
I know that $U_{\Gamma}=$Hom$(\Lambda, \Gamma)$ but don't I need $\bar{U_{\Gamma}}=$Hom$(\Lambda, \Gamma)$?
Thank you very much!
I have obtained the answer. Whenever Harris talks about the tangent space he means the Zariski tangent space as in the affine case. By this, his statement follows directly from the observation that the definition of the tangent space is local, i.e. if $X$ is a variety and $p\in U\subset X$ an open subset then $T_p(X)=T_p(U)$.
Sincerely, slin