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Let $X \subset \mathbb{C}^n$ be an affine variety (not irreducible). Let $Y$ be a subvariety of $X$ (again not irreducible). How can we relate the Zariski tangent space at $P \in Y$ and at $P \in X$?

(Corrected after Mariano's comments) Based on my understanding, we do have a homomorphism $T_P Y \rightarrow T_P X$ of vector spaces, but can we say something more? For example, what can we say about the dimensions of the two vector spaces $T_PY$ and $T_PX$?

Manos
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  • Do you mean "not irreducible" or "not necessarily irreducible"? – Nils Matthes Jul 20 '13 at 19:36
  • @NilsMatthes: Yes, i mean "not necessarily irreducible" :) – Manos Jul 20 '13 at 19:37
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    What homomorphism $T_pX\to T_pY$ do you have in mind? Notice that there are tons of such homomorphisms, simply because we could take the zero map, say, but presumably you have in mind a natural one. – Mariano Suárez-Álvarez Jul 20 '13 at 19:39
  • @MarianoSuárez-Alvarez: Yes you are right, i mean the canonical one. – Manos Jul 20 '13 at 19:41
  • and what is the canonical one?! – Mariano Suárez-Álvarez Jul 20 '13 at 19:42
  • @MarianoSuárez-Alvarez: It is the one induced by the canonical map between cotangent spaces $m_{P,X}/m^2_{P,X} \rightarrow m_{P,Y}/m^2_{P,Y}$. I am not sure though how it looks like :) – Manos Jul 20 '13 at 19:45
  • You are saying that that map induces a map $T_pX\to T_pY$? How? – Mariano Suárez-Álvarez Jul 20 '13 at 19:46
  • @MarianoSuárez-Alvarez: Since $\left(m_{P,X}/m^2_{P,X}\right)^* \cong T_PX$ by definition, where $*$ means dual, and each vector space of finite dimension is isomorphic to its dual, then we get a morphism $T_PX \rightarrow T_PY$. Right? – Manos Jul 20 '13 at 19:50
  • No, you do not get a morphism from $T_pX$ to $T_pY$ that way (and that the dual of a fin.dim. vector space is isomorphic to the vector space itself probably has nothing to do with what you want to say) – Mariano Suárez-Álvarez Jul 20 '13 at 19:51
  • @MarianoSuárez-Alvarez: Haha, ok, so...what should i say then? Is there another canonical or natural map between the tangent spaces? – Manos Jul 20 '13 at 19:54
  • Can you see why you do not get a map from $T_pX$ to $T_pY$ from your map $m_{p,X}/m_{p,X}^2\to m_{p,Y}/m_{p,Y}^2$ by taking duals? what do you get? – Mariano Suárez-Álvarez Jul 20 '13 at 19:56
  • @MarianoSuárez-Alvarez: No i can not see that, since if $f: m_{P,X}/m^2_{P,X} \rightarrow m_{P,Y}/m^2_{P,Y}$ is the canonical map of the cotangent spaces and we call $\psi_X : m_{P,X}/m^2_{P,X} \rightarrow T_PX$ the isomorphism between cotangent and tangent spaces, and similarly$\psi_Y : m_{P,Y}/m^2_{P,Y} \rightarrow T_PY$, then we certainly get a morphism $T_PX \rightarrow T_PY$ by $\psi_Y \circ f \circ \psi_X^{-1}$. Right? – Manos Jul 20 '13 at 20:05
  • Anything that involves an isomorphism between a vector space and its dual is wrong. – Mariano Suárez-Álvarez Jul 20 '13 at 20:06
  • If $f:V\to W$ is a linear map between vector spaces, what do you get by taking dual vector spaces? – Mariano Suárez-Álvarez Jul 20 '13 at 20:07
  • @MarianoSuárez-Alvarez: Ahh, i see, the order gets reversed...I will get $f^* : W^* \rightarrow V^*$. – Manos Jul 20 '13 at 20:14
  • There you go. ${}$ – Mariano Suárez-Álvarez Jul 20 '13 at 20:14
  • @MarianoSuárez-Alvarez: Thank you so much for teaching me!! – Manos Jul 20 '13 at 20:15
  • @MarianoSuárez-Alvarez: But could please explain the statement "anything that involves isomorphism between a vector space and its dual is wrong"? We know that if a vector space is finite dimensional, then it is isomorphic to its dual... – Manos Jul 20 '13 at 20:17
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    Dear @Manos, But the isomorphism is, in general, non-canonical, as it involves the choice of a basis. So if you're looking for a canonical map, choosing an isomorphism between a finite dimensional vector space and its dual is probably not going to help you. – Keenan Kidwell Jul 20 '13 at 21:10

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The natural $\Bbb{C}$-linear map $$ T_P(Y) \rightarrow T_P(X) $$ is indeed injective. This follows from the fact that it is dual to the $\Bbb{C}$-linear map $$ \mathfrak{m}_{X,P}/\mathfrak{m}_{X,P}^2 \rightarrow \mathfrak{m}_{Y,P}/\mathfrak{m}_{Y,P}^2 $$ which is surjective, since $Y$ is a subvariety of $X$. Hence one always has

$$ \dim T_P(Y) \leq \dim T_P(X) $$

and this result cannot be improved; you can have strict inequality (e.g. $X=\Bbb{A}^1$, $Y=P$ for some point $P \in \Bbb{A}^1$) and you can have equality (e.g. $X=\Bbb{A}^2$, $Y=V(y^2-x^3)$ and $P=(0,0)$).

Nils Matthes
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  • One more question please, to make sure i understand correctly the definitions: if all the gradients of the generators of the vanishing ideal of an affine variety $X$ of $\mathbb{C}^n$ are zero at a certain point $P$, then this means that the Zariski tangent space at this point is the entire $\mathbb{C}^n$. Right? – Manos Jul 20 '13 at 23:25
  • Yes, that's right, Manos. – Georges Elencwajg Jul 21 '13 at 06:00