How to see the second fundamental form is symmetric when defining it as the derivative of the Gauss map?

Question

Let $X \subset \mathbb C^N$ be an $n$-dimensional complex submanifold. Voisin writes in [1]

Let us introduce the second fundamental form $$\Phi: S^2 T_{X,x} \to \mathbb C^N / T_{X,x}$$ which can be defined as the differential of the Gauss map $$\varphi:X \to \operatorname{Grass}(n,N),\ x \mapsto T_{X,x}.$$

Set $G = \operatorname{Grass}(n,N)$, and let $0 \to A \to \underline{\mathbb C}^n \to B \to 0$ be the universal short exact sequence of vector bundles on $G$. The differential of the Gauss map fits into a diagram $$\require{AMScd} \begin{CD} T_X @>>> T_G @= \operatorname{Hom}(A,B) \\ @VVV @VVV @.\\ X @>>> G @. \end{CD}$$ As $A_{\varphi(x)} = T_{X,x}$ and $B_{\varphi(x)} = \mathbb C^N / T_{X,x}$, we get bilinear maps $$T_{X,x} \times T_{X,x} \to \mathbb C^N / T_{X,x}.$$

But why is this map symmetric?

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[1] Voisin, Hodge Theory and Complex Algebraic Geometry II

Answer 1

$\newcommand\C{\mathbb{C}}$UPSHOT: If $f: X \rightarrow \C^N$ is the inclusion map, $\partial^2f$ is the Hessian with respect to local coordinates, and $\pi_x: \C^N \rightarrow \C^N\backslash T_xX$ is the quotient map, then it is easy to show that, $$ x \mapsto \pi_x(\partial^2 f(x)) $$ is a section of $S^2T^*X\otimes \C^N/T_*X$. The longwinded discussion below is an attempt to explain why it is the second fundamental form.

ORIGINAL ANSWER:

As always, symmetry like this arises from the fact that partials commute.

Here's one way to do it using local coordinates and affine coordinates for the Grassmannian:

Given $p \in X$, move $X$ to a position such $p$ is at the origin, and the tangent plane at $p$ is $T_0 = \{z^{n+1}=\cdots =z^N = 0\}$. In a suitably chosen neighborhood $O \subset \C^N$ of $0$, $X$ is a graph of a map $f: T_0 \rightarrow \C^{N-n}$, i.e., $$ X = \{ (z,f(z))\}. $$

The Grassmannian $G(n,N)$ in a neighborhood $N$ of $T_0$ is parameterized by linear maps $L: T_0 \rightarrow \C^N/T_0$. Each linear map maps to its graph. In particular, if $\mathcal{L}$ denotes the space of all linear maps from $T_0$ to $\C^N/T_0$, then there is a natural injective map \begin{align*} \mathcal{L} &\rightarrow G(n,N)\\ L &\mapsto \{ v + L(v):\ v \in T_0\}. \end{align*}

The Gauss map is the map from $z\in$ near $0$ to $$ \phi(z) = T_zX = \{ v + \partial f(z)(v):\ v \in T_0 \} \in G(n,N). $$ Equivalently, using the affine coordinate map $\mathcal{L} \rightarrow G(n,N)$, it defines a map \begin{align*} X &\rightarrow \mathcal{L}\\ z &\mapsto \partial f(x). \end{align*} Since $\partial f(x)$ is a linear map from $T_0$ to $\C^N/T_0$, its derivative at $0$ is a map $$ \partial^2f(0): T_0\otimes T_0 \rightarrow \C^N/T_0. $$ Since partial commute, it is really a map $$ \partial^2f(0): S^2T_0 \rightarrow \C^N/T_0. $$ This is the second fundamental form.

There are, of course, details to check, including verifying that this really is the second fundamental form and that it is a tensor (i.e, a bundle map). What is crucial is that the second fundamental form at each point $x \in X$ is obtained by writing $X$ locally as a graph over $T_xX$ of a function $f: T_xX \rightarrow \C^N/T_xX$. Then $0 \in T_xX$ is a crticial point of $f$, and, as a consequence, the Hessian of $f$ at $0$ is a well-defined tensor.

Answer 2

Those are my computational details to Deane's answer:

We have wlog $x = 0 \in X$, and $V = T_{X,x} = \langle e_1, \dotsc, e_n \rangle_{\mathbb C} \subset \mathbb C^N$. Let $D \subset \mathbb C^n$ be a small disc, and let $f: D \to \mathbb C^N$ be a local parametrisation of $X$, with $f(0) = 0$. For $z \in D$, the tangent space $T_{X,f(z)}$ is generated by the columns of the differential matrix $$Df = \begin{pmatrix} \frac{\partial f_1}{\partial z_1}(z) & \dots & \frac{\partial f_1}{\partial z_n}(z) \\ \vdots && \vdots \\ \frac{\partial f_N}{\partial z_1}(z) & \dots & \frac{\partial f_N}{\partial z_n}(z) \end{pmatrix}.$$ As Deane explained, we have an open inclusion $\operatorname{Hom}(V,W) \hookrightarrow G(n,N)$, which maps a map $\varphi: V \to W$ to the subspace $(\operatorname{id} + \varphi)(V).$ Now with respect to the two open charts chosen, the Gauß map $X \to G(n,N), x \mapsto T_{X,x}$ is given by cutting the upper rows of $Df$ ("projecting to $W$"), i.e. $$D \ni z \mapsto \begin{pmatrix} \frac{\partial f_{n+1}}{\partial z_1}(z) & \dots & \frac{\partial f_{n+1}}{\partial z_n}(z) \\ \vdots && \vdots \\ \frac{\partial f_N}{\partial z_1}(z) & \dots & \frac{\partial f_N}{\partial z_n}(z) \end{pmatrix} \in \operatorname{Hom}(V,W).$$ Differentiating this yields the Hessian matrices for the $f_i, i > n$.