Differential of a smooth map computation

Question

I wanted to compute the differential of the map $f:M_n(\mathbb{R})\to S_n(\mathbb{R})$ defined by $f(A)=AA^T$ but how can I compute it explicitly?

it has been done here but I am not sure how and why it is being done that way.

What I understand is, given a map $F:M\to N$, the differential $DF_p:T_pM \to T_qN$ is the differential map. Now the differentials are nothing but special functions $\omega:C^\infty(M)\to \mathbb{R}$ satisfying some properties. Now let us focus on the differential map $ DF_p:T_pM \to T_qN$, for a given $\omega:C^\infty(M) \to \mathbb{R}$, we need to define a map $\omega': C^\infty(N) \to \mathbb{R}$ using $\omega$. But how are we going to compute it explicitly using curves?

I also know that any such differential $\omega$ is given by a smooth curve $\gamma_\omega:I \to M$ as follows

$$\omega(g)=\frac{d( g \circ \gamma_\omega)}{dt}$$ but I am not able to understand how to compute the differentials

Answer 1

The question you linked computed the directional derivative call it $f^{\text{directional derivative}}(p,v)$ of $f$ at any point $p$ and in any direction $v$, as you have agreed in the comments. Now, to define $df_p$, we can simply evaluate

$$df_p(v) = f^{\text{directional derivative}}(p,v)$$

So, in this case, for your $f$,

$$df_p(v) = v^tp+p^tv $$

due to the logic written in the answer you have linked. Basically your question amounts to how exactly the derivative and directional derivatives are linked in finite dimensions, since you are using an answer that gives the directional derivatives for your $f$.

Answer 2

For any manifold $M$, you can consider the tangent space at a point $p\in M$ as the set of curves through that point modulo the equivalence relation $\gamma_1\sim\gamma_2 \iff \gamma_1(0)=p =\gamma_2(0)\ \text{and}\ \gamma_1'(p)=\gamma_2'(p)$. In this setting, $DF_p(v)$ for $v\in T_pM$ is computed as follows:

1. Pick a curve $\gamma\in M$ such that $\gamma'(0)=v\in T_pM$.
2. Compute $$ DF_p(v)=\frac{d}{dt}\big(F\circ \gamma\big)\big\vert_{\ t=0} $$

If you think about it for a second, you will see that this is nothing but computing the velocity of the image of $\gamma$ under $F$ at $F(p)$. In some sense, you are pushing forward $v$ via $\gamma$.

As for your example, observe that $M_n(\mathbb{R})\cong \mathbb{R}^{n^2}$ and $S_n(\mathbb{R})\cong \mathbb{R}^{n(n+1)/2}$. Since they are Euclidean, we can identify them with their tangent spaces, i.e. $T_pM_n(\mathbb{R})\cong M_n(\mathbb{R})$ and $T_{F(p)}S_n(\mathbb{R}) \cong \mathbb{R}^{n(n+1)/2}$. Hence the differential is a map of the following form $$DF_p: M_n(\mathbb{R})\longrightarrow S_n(\mathbb{R})$$ Now for $v\in T_pM_n(\mathbb{R})$, which we know is just an $n\times n$ matrix, we can simply pick the curve $$\gamma(t) = p + tv$$ It is clear that $\gamma$ satisfies $\gamma(0)=p$ and $\gamma'(0)=v$. Now by step $(2)$ above, we have \begin{align} DF_p(v) &= \frac{d}{dt}\big(F\circ \gamma\big)\big\vert_{\ t=0} = \frac{d}{dt}\big(F(p+tv)\big)\big\vert_{\ t=0} = \frac{d}{dt}\big((p+tv)(p+tv)^T\big)\big\vert_{\ t=0} \\ &=\frac{d}{dt}\big((p+tv)(p^T+tv^T)\big)\big\vert_{\ t=0} = \frac{d}{dt}\big(pp^T+tpv^T+tvp^T+t^2vv^T\big)\big\vert_{\ t=0} \\ &= pv^T+vp^T \end{align} as promised.

Answer 3

We have

\begin{align} f(A+H) &= (A+H)(A+H)^T = (A+H)(A^T+H^T)\\[1.5ex] &= AA^T+AH^T+HA^T+HH^T\\[1.5ex] &= f(A) + AH^T+HA^T+HH^T \end{align}

so

$$ df(A)(H) = AH^T+HA^T $$

being $\,H\longmapsto AH^T+HA^T\,$ a linear continuous function $M_n(\mathbb R)\longrightarrow S_n(\mathbb R)$, $\,$and $\,HH^T = o(H)$.

Note that the directional derivative of $f$ at $A$ along $H$ is given by

$$ D_Hf(A) = df(A)(H).$$