# Tangent bundle and normal bundle in self-product

$\newcommand{\I}{\mathcal{I}}$ Let $X$ a variety smooth over the complex numbers. Then we know that $\Omega_{X/\mathbb{C}}$ is the (usual) pullback of the conormal sheaf $\I/\I^2$ where $\I$ the sheaf of ideals of the diagonal in $X\times_\mathbb{C} X$. You can of course prove this directly.

One could expect something of this nature (taking $\Omega_X$ to be defined only as the sheaf of Kahler differentials) from computations with the conormal exact sequence

$$0 \rightarrow \I /\I^2 \rightarrow \Omega_{X\times X} \otimes \mathcal{O}_\Delta \rightarrow \Omega_\Delta \rightarrow 0$$

$\textbf{Question:}$ Are there geometric ways to see this also? Is there a picture that geometers keep in mind making it obvious that say the tangent bundle to $X$ and the normal bundle of the diagonal $\Delta \subset X\times X$ are really the same thing? This is a very basic concept, yet I can't find such an explanation. For example, is this clear in the context of differential geometry or complex manifolds?

• Morally, this is actually the way we introduce vectors in (pre)calculus or in school, right? We talk of pairs of points $P$,$Q$ in $\mathbb{R}^2$ (up to equivalence), the "diagonal pair" $(P,P)$ being (equivalent to) zero. Commented Nov 4, 2012 at 9:20
• Peter, I'm not sure I understand what you are saying. Could you elaborate a little?
– LMN
Commented Nov 4, 2012 at 14:13
• I'll try. Since a smooth projective variety is locally a manifold, it locally looks like affine $\mathbb R^n$, so $X \times X$ looks like affine $\mathbb R^n \times \mathbb R^n$, and the diagonal looks like the diagonal. The normal bundle consists of points in affine $\mathbb R^n \times \mathbb R^n$, up to translation by the diagonal, in other words, it consists of differences of points in $\mathbb R^n$. As those differences get very small they look like tangent vectors. Commented Nov 4, 2012 at 15:48
• @Will, @LMN: Exactly! A vector $PQ$ in $\mathbb{R}^n$ has a "tail" (P) and "head" (Q). A (smooth) vector field on $\mathbb{R}^n$ can be thought of as a map $\mathbb{R}^n\to\mathbb{R}^n\times\mathbb{R}^n$, $P\mapsto (P,Q)$. And $P\mapsto (P,P)$ is the zero section, $PQ=OQ-OP$. To make sense of this properly you need the tangent sequence of the diagonal, see Michael's or Sandor's answer. I am saying something really trivial, though. Commented Nov 4, 2012 at 16:32

## 2 Answers

In the case of differential geometry everything reduces to vector spaces. Let $x \in X$. Then at any point $(x, y) \in X \times X$ $$T_{(x, y)} X \times X = T_x X \oplus T_y X .$$ Using this identification the tangent to the diagonal at a point $(x, x)$ is the subspace of $T_{(x, x)} X \times X$ given by $$T_{(x, x)} (\Delta) = \lbrace (\xi, \xi ) \mid \xi \in T_x X \rbrace \subset T_{(x, x)} X \times X.$$ On the other hand the normal is the quotient $$(T_{(x, x)} X \times X) / T_{(x, x)} \Delta$$ and we can identify this with $T_x X$ in at least two slightly different ways. Either $$\iota_1 \colon \xi \mapsto (\xi , - \xi) + T_{(x, x)} \Delta$$ or $$\iota_2 \colon \xi \mapsto (-\xi , \xi) + T_{(x, x)} \Delta .$$

We can of course also identify $T_x X$ and $T_{(x, x)} \Delta$ by $\xi \mapsto (\xi, \xi)$.

I guess one explanation for the two identifications of the normal bundle is that there is involution $\tau \colon X \times X \to X \times X$ given by $\tau(x, y) = (y, x)$ which fixes the diagonal pointwise and hence acts trivially on the tangent space to the diagonal. As a result it descends to an action on the normal bundle which interchanges the two identifications $\iota_1$ and $\iota_2$, that is $\tau \circ \iota_1 = \iota_2$

• I think this explanation is given in Milnor and Stasheff. Commented Nov 4, 2012 at 8:00
• That would make sense. Probably where I learnt it. I'll check my copy in the office tomorrow. Commented Nov 4, 2012 at 12:05
• @Michael, Sándor, so do I understand correctly, the reason for the identification is because we have short exact sequences relating various tangent and normal bundles (really just the conormal exact sequence stated above) and like you (both) show, we can explicitly find that the tangent bundle and the normal bundle of the diagonal are the same bundle on X. My point (and question) is that this is the way we see that there is such an identification of bundles, not because it is "geometrically obvious" in some other way.
– LMN
Commented Nov 4, 2012 at 15:22
• @LMN: I added a bit to my answer to respond to this. Commented Nov 4, 2012 at 19:34

Let $p_1,p_2:X\times X\to X$ be the two projections to $X$. Then $\Omega_{X\times X}\simeq p_1^*\Omega_X\oplus p_2^*\Omega_X$ and the residue map $\Omega_{X\times X}\otimes \mathscr O_{\Delta}\to \Omega_\Delta$ induces an isomorphism $p_1^*\Omega_X\otimes \mathscr O_{\Delta}\simeq \Omega_\Delta$, since $p_1|_\Delta$ is an isomorphism.Then it is easy to see that the kernel of this map is isomorphic to $p_2^*\Omega_X\otimes \mathscr O_{\Delta}\simeq \Omega_\Delta$.

So, does this mean that it should be geometrically intuitive that indeed the co-normal bundle of the diagonal is isomorphic to the cotangent sheaf? I say yes: If you repeat this argument in plain words it says that the (co)tangent sheaf of $X\times X$ restricted to the diagonal is naturally (cf. what does "natural" mean?) the direct sum of two copies of the (co)tangent sheaf of $\Delta\simeq X$. Furthermore, when we decompose the restriction of (co)tangent vectors of $X\times X$ to (co)tangent and (co)normal vectors of $\Delta$, then since the (co)tangent of $\Delta$ can be identified with one of the two components of the (co)tangent of $X\times X$, the (co)normal gets identified with the other.

I would add one more important issue that can get lost both in an intuitive and a local argument. It is important to keep track of what maps where naturally. Verifying an isomorphism of sheaves locally only works if one already has a morphism globally. In particular, in Michael's answer, when he says "...the normal is the quotient...", he is using the fact that there is such a map globally that restricts to this one. Otherwise one would only prove that there is a local isomorphism, which between locally free sheaves only means that they have the same rank. In other words, to do this correctly you need to argue with the (global) short exact sequence. Then again, if you truly understand that sequence, it is equivalent to the intuitive argument above about (co)tangents.