# Second fundamental form for Lie group $SO(n)$

I'm given $$\mathbb{R}^{n^2} = \mathbb{R}^{n \times n}$$ with the usual metric $$\langle A, B \rangle = \text{Trace} (AB^T) = \text{Trace} (A^T B)$$.

Then $$SO(n)$$ is a submanifold of $$\mathbb{R}^{n^2}$$ of dimension $$\frac{1}{2} n (n-1)$$.

I have shown that the tangent space $$T_{I} SO(n)$$ consists of all skew-symmetric matrices with zero trace.

I wish to show that the normal space $$N_I SO(n)$$ consists of symmetric matrices. Furthermore, I need to calculate the second fundamental form tensor for $$SO(n)$$.

I'm given $$X$$ and $$Y$$ left-invariant vector fields determined by skew-symmetric matrices at the identity. If $$O$$ is an arbitrary orthogonal matrix, then $$X(O) = OA$$ and $$Y(O) = OB$$.

I need to prove that $$D_Y X(O) = \frac{1}{2} O [B, A] + \frac{1}{2} O (BA + AB)$$ and to find expressions for $$\nabla_Y X$$, $$h(X,Y)$$ and $$[X, Y]$$.

Here $$\nabla_X Y$$ denotes the covariant derivative on $$SO(n)$$, while $$D_X Y$$ denotes the covariant derivative on $$\mathbb{R}^{n^2}$$. Also, $$h(X,Y)$$ denotes the second fundamental form tensor.

Attempt:

I know that $$D_X Y = \nabla_X Y + h(X,Y)$$ is the Gauss formula.

On $$\mathbb{R}^{n^2}$$, I have $$D_Y X (O) = D_{Y(O)} X = Y(O) X.$$ To calculate the latter, we take a curve through $$O$$ with tangent vector $$Y(O) = OB$$. Such a curve is $$O \exp(Bt)$$. Then $$Y(O) X = \frac{d}{dt}_{t = 0} X (O \exp (Bt)) = OBA.$$

How can I prove that $$2 h(X,Y) = D_X Y + D_Y X$$ ? This would give me that $$h(X,Y) (O) = \frac{1}{2} O (AB + BA).$$

Also, I'm not sure how to calculate $$\nabla_Y X$$.

It seems to me that you've done roughly half of the work, and the part you're missing is seeing why $$\nabla_XY|_O = \frac{1}{2}O[X,Y]$$. This actually is a general result about Lie groups equipped with a bi-invariant metric (a metric which is both left- and right-invariant). A nice result that's often cited is that compact Lie groups are guaranteed a bi-invariant metric.

Proposition: If $$G$$ is a Lie group with a bi-invariant Riemannian metric $$\langle\cdot,\cdot\rangle$$, then it satisfies: $$\langle [X,Y], Z\rangle \;\; =\;\; - \langle Y, [X,Z]\rangle.$$

Proof: Letting $$X,Y,Z \in \mathfrak{g}$$ be left-invariant vector fields, we can compute $$\left .\frac{d}{dt} \langle Ad_{\gamma(t)}Y, Ad_{\gamma(t)}Z\rangle \right |_{t=0}$$, where $$\gamma(t) = \exp(tX)$$ and $$Ad_{\gamma(t)} = \left (L_{\gamma(t)}\right )_*\circ \left (R_{\gamma(-t)}\right)_*$$. Notice here that the bi-invariance of the metric yields

$$\left . \frac{d}{dt} \langle Ad_{\gamma(t)}Y, Ad_{\gamma(t)}Z \rangle \right |_{t=0} \;\; =\;\; \left . \frac{d}{dt} \langle Y,Z\rangle \right |_{t=0} \;\; = \;\; 0$$

where $$\langle Y,Z\rangle$$ is constant due to these being left-invariant vector fields. Observe on the other hand that $$\begin{eqnarray*} \left . \frac{d}{dt} \langle Ad_{\gamma(t)}Y, Ad_{\gamma(t)}Z\rangle \right |_{t=0} & = & \langle XY-YX, Z\rangle + \langle Y, XZ - ZX \rangle \\ & = & \langle [X,Y], Z\rangle + \langle Y, [X,Z]\rangle \;\; = \;\; 0. \end{eqnarray*}$$ $$\tag*{\Box}$$

Proposition: For a bi-invariant Lie group $$G$$, the covariant derivative is given by
$$\nabla_XY \;\; =\;\;\frac{1}{2}[X,Y]$$ for left-invariant vector fields $$X,Y \in\mathfrak{g}$$. Evaluated at a particular point on the Lie group, this expression is given by $$\left . \nabla_XY\right |_Q =\frac{1}{2}Q[X,Y]$$.

Proof: A general result from Riemannian geometry is that a Riemannian metric $$\langle \cdot, \cdot\rangle$$ satisfies the equation $$\langle\nabla_XY,Z\rangle \;\; =\;\; \frac{1}{2} \left (X\langle Y,Z\rangle + Y\langle Z,X\rangle - Z\langle X,Y\rangle - \langle Y, [X,Z]\rangle - \langle Z,[Y,X]\rangle + \langle X, [Z,Y]\rangle \right ).$$

For each term of the form $$X\langle Y,Z\rangle$$, the action of $$X$$ is that of taking a derivative of the inner product $$\langle Y,Z\rangle$$ at each point. In the case of a bi-invariant Lie group with left-invariant vector fields $$X,Y,Z \in \mathfrak{g}$$, these terms all vanish, hence our equation in our special case reduces to

$$\langle \nabla_XY, Z\rangle \;\; =\;\; \frac{1}{2} \left (-\langle Y, [X,Z]\rangle - \langle Z, [Y,X]\rangle - \langle X, [Z,Y]\rangle \right ).$$

By the first proposition we have the first and third terms both vanish, and we're left with

$$\langle \nabla_XY, Z\rangle \;\; =\;\; -\frac{1}{2}\langle Z, [Y,X]\rangle \;\; =\;\; \left \langle Z, \frac{1}{2}[X,Y]\right \rangle \;\; =\;\; \left \langle \frac{1}{2}[X,Y], Z\right \rangle$$

where we used the anti-symmetry of the bracket and the symmetry of the metric. Since this equation is true for all left-invariant vector fields we conclude that $$\nabla_XY = \frac{1}{2}[X,Y]$$ on the left-invariant vector fields of $$G$$.
$$\tag*{\Box}$$

If we combine these two results with your computation that $$D_XY(O) = OAB$$, then we take the second-fundamental form $$D_XY = \nabla_XY + h(X,Y)$$. Noting the skew-symmetry of the bracket we have that $$\begin{eqnarray*} D_XY + D_YX & = & \nabla_XY + \nabla_YX + 2h(X,Y) \\ & = & \frac{1}{2}O[X,Y] + \frac{1}{2}O[Y,X] + 2h(X,Y) \\ & = & \frac{1}{2}O[X,Y] - \frac{1}{2}O[X,Y] + 2h(X,Y) \\ & = & 2h(X,Y). \end{eqnarray*}$$

From here you can conclude that $$h(X,Y) \;\; =\;\; \frac{1}{2}O(XY + YX).$$

Note: A subtle point here is the fact that we assume $$h(X,Y) = h(Y,X)$$ even though in general the second fundamental form isn't necessarily a symmetric tensor. The reason why this works here is precisely because $$N_ISO(n)$$ consists of symmetric matrices, and by extension we have that $$N_QSO(n) = \left \{QS \; | \; S = S^T\right \}$$. This fact can actually be proven just with results about matrices and the trace. Since you've successfully proven that $$T_ISO(n)$$ is the space of skew-symmetric matrices, try showing that if $$A \in T_ISO(n)$$ then $$Tr\left (A^TB\right ) = 0$$ if and only if $$B = B^T$$.