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Question: Consider the smooth map $$F : M(n , \mathbb{R} ) = \mathbb{R}^{{n}^{2}} \to S(n,\mathbb{R}) = \mathbb{R}^{\frac{n(n+1)}{2}} \hspace{1cm} A \mapsto AA^{T}$$ As the differential $$dF_A : T_A M(n,\mathbb{R}) = M(n,\mathbb{R}) \to T_{I}S(n,\mathbb{R}) = S(n,\mathbb{R}).$$ Show that for every $A \in O(n)$, $$T_A O(n) = \ker dF_A = \left\{ B \in M(n,\mathbb{R}) : A^{T}B + B^{T}A = 0 \right\} $$

Notation: $$M(n, \mathbb{R}) = \left\{ A \space\ | \space\ A \space\ \text{is an} \space\ n \times n \space\ \text{matrix} \right\}$$ $$S(n, \mathbb{R}) = \left\{ A = A^{T} \space\ | \space\ A \in M(n, \mathbb{R}) \right\}$$ $$O(n) = \left\{ A^{-1} = A^{T} \space\ | \space\ A \in M(n, \mathbb{R}) \right\}$$

My ideas so far:

I have shown that $dF_A$ is onto for every $A \in O(n) = F^{-1}(I)$, and that the orthogonal group $O(n)$ is a submanifold of $M(n, \mathbb{R})$ of codimension $\frac{n(n+1)}{2}$.

Did
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Dragonite
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  • "My ideas so far" ?? Why not simply compute $dF_A$ (its kernel then becoming quite apparent)? By the way, what is the definition of $dF_A$? – Did Feb 27 '17 at 15:28
  • The definition I have is that for if $M$ and $N$ are smooth manifolds and $F : M \to N$ is a smooth map, for each $p \in M$, we define a map $dF_p : T_p M \to T_{F(p)} N $ called the differential. So in this context, $M = M(n, \mathbb{R})$, and $N = S(n, \mathbb{R})$ and $F$ is such that $A \mapsto AA^{T}$. I don't understand how to go about computing $dF_A$. – Dragonite Feb 27 '17 at 15:42
  • Look at the definition of $dF_A$ then (bis). – Did Feb 27 '17 at 15:44
  • From my understanding the definition of the differential is like a generalization of the Jacobian - so I don't know how to go about computing it in this context. – Dragonite Feb 27 '17 at 15:50
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    What? IOW, you have no definition of $dF_A$ and you are trying to solve this exercise? Sorry but this does not make any sense (and the bounty, oh my...). For starters, see http://math.stackexchange.com/a/132417/ – Did Feb 27 '17 at 15:54

1 Answers1

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I believe I got most of it.

Let $\gamma$ be a curve with its target space being $M(n, \mathbb{R})$, say, $\gamma : J \to M(n,\mathbb{R})$ such that $0 \in J \subseteq \mathbb{R}$, where $\gamma(t) = tB + A$ so that $\gamma(0) = A$, and $\gamma'(0) = B$. Then, we have that $dF_A (B) = (F \circ \gamma)'(0)$. \begin{align*} (F \circ \gamma)(t) &= F(\gamma(t)) \\ &= F(tB + A) \\ &= (tB + A)(tB + A)^{T} \\ &= (tB + A)(tB^{T} + A^{T}) \\ &= t^{2}BB^{T} + tBA^{T} + tAB^{T} + AA^{T} \end{align*} Taking this derivative we find that, $(F \circ \gamma)'(t) = 2tBB^{T} + BA^{T} + AB^{T}$, and then evaluating at $t = 0$, we have that $dF_A (B) = (F \circ \gamma)'(0) = BA^{T} + AB^{T}$. Then, we know that the kernel of $dF_A$ are those $B \in M(n,\mathbb{R})$ such that $BA^{T} + AB^{T} = 0$, and hence we have that $$\ker dF_A = \left\{ B \in M(n,\mathbb{R}) : BA^{T} + AB^{T} = 0 \right\} $$ However, this is a little different than what we aim to show. But, if $A \in O(n)$ so that $A^{T} = A^{-1}$, so that $AA^{T} = A^{T}A = I$ and working with the equation \begin{align*} BA^{T} + AB^{T} = 0 &\iff A^{T}BA^{T} + A^{T}AB^{T} = 0 \\ &\iff A^{T}BA^{T} + B^{T} = 0 \\ &\iff A^{T}BA^{T}A + B^{T}A = 0 \\ &\iff A^{T}B + B^{T}A = 0 \end{align*} so that for $A \in O(n)$ these two sets are equivalent. Hence, $$T_A O(n) = \ker dF_A = \left\{ B \in M(n,\mathbb{R}) : A^{T}B + B^{T}A = 0 \right\} $$

Dragonite
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