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Let $f:SO(n)\rightarrow S^{n-1}$, $f(A)=(A^n_i)_i$, that is $f(A)$ is the last row of $A$. Show that $f$ is a submersion.

I'm not sure how to calculate $df$, because I only know how to calculate the differential using local charts, but I don't know how to parametrize $SO(n)$, so this is my attempt:

Let $F:M_n(\mathbb{R})\rightarrow\mathbb{R}^n$, $F(M)=(M^n_i)_i$. Then $F|SO(n)=f$. Since $F$ is linear, if $p\in SO(n)\subset M_n(\mathbb{R})$, $dF_p(v)=(v^n_i)_i$, and we conclude that $df_p(v)=(v^n_i)_i.$

So, this is right? And how can I show that df is surjective?

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Firstly, remark that if $I$ is the identity of $SO(n), T_ISO(n)=AS(n)$ the space of skew symmetric matrices such that for every $A\in T_ISO(n), A^T+A=0$, $df_I$ is surjective. To see this remark that for every $(v_1,v_2,...,v_{n-1})$, there exists a skew symmetric matrix $(b_{ij})$ whose last column verifies, $b_{in}=v_i, b_{nn}=0$, $df_I((b_{ij})=(v_1,...,v_{n-1},0)$. This implies that $df_I$ is surjective.

Let $g$ be any element of $SO(n)$, $T_gSO(n)=gT_ISO(n)$, this implies that for every skew-symmetric matrix $A$, $df_g(gA)=gdf_I(A)$ is the last column of $gA$, since $g$ is invertible, and the image of $df_I$ has dimension $n-1$, the image of $df_g$ has dimension $n-1$ since it is the image of $df_I$ by $g$, henceforth $g$ is a submersion.

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  • $\begingroup$ I didn't understand why $df_g(gA)=gdf_I(A)$. In $gdf_I(A)$ you are thinking of $g$ as a linear transformation on $T_{f(I)}S^{n-1}$? $\endgroup$ – Fernando Escobosa Apr 24 '16 at 15:31
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In fact, if $e_n$ is the $n-$th base vector $f(A)= Ae_n$. Let us extend this formula to the set of all matrices, $F(A)=A.e_n$. So $F$ is linear and is its own differential $F'(A)B= Be_n$. The derivative of $f$ at the point $A$ is the restriction of this map to the tangent space $ \cal {A}(n). A$ of $O(n)$ at $A$, where $\cal A$ is the set of antisymmetric matrices. So we are reduce to prove that for every vector $y$ perpendicular to $Ae_n =u$ we can find an antisymmetric matrix $B$ such that $y=B (Ae_n)=Bu$. But precisely ${\cal A} .u=u^{\perp}$, as one can check in an orthonormal basis where $u$ is the first vector.

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