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Define a smooth vector field $X_S$ on the orthogonal group, $O(n)$, (group of matrices such that $A^{T} = A^{-1}$), where $X_S (A) = SA$, where $S$ belongs to the space of skew symmetric matrices ( $S^{T} = - S$). Why is $X_S$ left invariant?

I know that we say a smooth vector field $X:G \to TG$ is left invariant if $$dL_g\circ X=X\circ L_g, \ \forall g\in G \space\ \space\ \text{where} \space\ \space\ L_g(h):=gh.$$ I also see that $$T_A O(n) = \left\{ B \in M(n,\mathbb{R}) \space\ \bigg| \space\ BA^{T} + AB^{T} = 0 \right\}$$ I think I'm misunderstanding the computation here (the notation confuses me), \begin{align*} dL_g \circ X_S (A) &= dL_g(SA) \\ &= \\ &\vdots \\ &=X_S \circ L_g (A) \end{align*}

Dragonite
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1 Answers1

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Take a look in my counterexample. Choose $n=2$, $S= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $ and $G= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $. So we have $L_G(A)=GA \implies d(L_G)_A(B)=GB $ where $B \in T_A O(n)$.

Therefore, $d(L_G)_A(X_S(A))=GX_S(A)=GSA$.

Choose $A= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. So $d(L_G)_A(X_S(A)) = GS= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$.

But $X_S(L_G(A))=X_S(GA)=SGA=SG= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$. Therefore, $X_S$ is not left invariant.

Danuso Rocha
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  • $L_G(A)=GA \implies d(L_G)_A(B)=GB $ where $B \in T_A O(n)$

    Why does that imply $d(L_G)_A(B) = GB$, and then in your next line wouldn't that mean $X_S (A) \in T_A O(n)$ too?

     – Dragonite Mar 20 '17 at 14:32
  • Maybe this notation is a little confusing, I agree. But change the notation to $f:=L_G$. So we have $d(f)_A(B)=GB$. – Danuso Rocha Mar 20 '17 at 14:36
  • What's the difference between $d(L_G)_A$ and $L_G (A)$? It seems like they are both just left multiplication by $G$ from what you wrote? Which is what $L_G (A)$ is, I know. But what about $d(L_G)_A$, again it's lack of understanding of notation, sorry for the elementary questions. – Dragonite Mar 20 '17 at 14:43
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    First $X_S(A) \in T_AO(n)$, because $X_S(A)A^T+ A(X_S(A))^T = SAA^T+A(SA)^T = S + A(A^TS^T)=S+S^T=0$. – Danuso Rocha Mar 20 '17 at 14:52
  • Ahh, that makes perfect sense for why $X_S (A) \in T_A O(n)$. – Dragonite Mar 20 '17 at 14:55
  • take a look at "The differential of a smooth map" here:https://en.wikipedia.org/wiki/Pushforward_(differential) . In our case, $L_G=\phi$ and $A=x$, so I took the differential of $L_G$ at the point $A$, that is, $d(L_G)_A$. – Danuso Rocha Mar 20 '17 at 14:57