Denoting the components of the $3\times3$ matrix $A \in O(3)$ as $a_{ij}$, show that

$$ F: O(3) \rightarrow S^2, a_{ij} \mapsto a_{1j} $$

is a submersion. (The map is well defined since for $A \in O(3)$ it is true that $a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$.)

From what I understand to show that the map is a submersion, I have to show that the differential map

$$ dF: T_AO(3) \rightarrow T_{F(A)}S^2 $$

is onto for every $A \in O(3)$.

Now if I were given a tangent vector $X\in T_AO(3)$ then from what I understand the map simply is $dF: x_{ij} \mapsto x_{1j}$, where $x_{ij}$ denote the components of $X$. However, I fail to see that this is surjective at every point $A\in O(3)$.

Indeed at the identity $A = I$ I know that a basis of tangent vectors is given by the antisymmetric matrices with one non-zero number in the upper triangular part and that then the map is something like $X \mapsto (0, s, t)$ and arguably two parameters are enough to span the tangent space of $S^2$.

I fail to see how this is true for tangent vectors $X$ at an arbitrary point $A$ with the vector being given by

$$ X = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \gamma(t)$$

where $\gamma: \mathbb{R} \rightarrow O(3)$ with $\gamma(0) = A$.


You can see $O(3)\subset S^2\times S^2\times S^2\subset \mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3$ as the set of ordered orthonormal basis of $\mathbb{R}^3$, a matrix $A$ being interpreted as three aligned vectors $\pi_1(A),\pi_2(A)$ and $\pi_3(A)$, where $\pi_i:\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R}^3$ is the projection onto the $i$-th factor. So your map $F$ is the restriction $\pi_1|_{O(3)}$.

A (smooth) path $\gamma:\mathbb{R}\to S^2$ being fixed, with $\gamma=F(A)=\pi_1(A)$ for $A\in O(3)$, you would like to find a path $\Gamma=(\gamma_1,\gamma_2,\gamma_3):\mathbb{R}\to O(3)$ such that $\Gamma(0)=A$ and $\pi_1\circ\Gamma=\gamma$, i.e. $\gamma_1=\gamma$. So your question rephrases as: is it always possible to complete a path of vectors of norm $1$ in a path of orthonormal basis? The answer is yes, by Gram–Schmidt process.

More formally, fix a path $\gamma:(-\varepsilon,\varepsilon)\to S^2$ such that $\gamma(0)=F(A)=\pi_1(A)$ and $\gamma'(0)=E$ for some $A\in O(3)$ and an arbitrary $E\in T_{F(A)}S^2$. We will complete it in a path $$\Gamma=(\gamma,\gamma_2,\gamma_3):(-\varepsilon,\varepsilon)\to S^2\times \mathbb{R}^3\times\mathbb{R}^3$$ such that forall $t\in(-\varepsilon,\varepsilon),\Gamma(t)=(\gamma(t),\gamma_2(t),\gamma_2(t))$ is an orthonormal base of $\mathbb{R}^3$:

  • We choose $\gamma_2(t)\equiv \pi_2(A)$ and $\gamma_3(t)\equiv \pi_3(A)$ (note that $\Gamma(0)=(\gamma(0),\gamma_2(0),\gamma_3(0))=(\pi_1(A),\pi_2(A),\pi_3(A))=A\in O(3)$ is an orthonormal base of $\mathbb{R}^3$).

  • By continuity of the determinant, $(\gamma(t),\gamma_2(t),\gamma_3(t))$ is still a base of $\mathbb{R}^3$ for $t$ small enough: restrict the interval in order to keep this true on all of it.

  • Then apply the Gram–Schmidt process on $(\gamma(t),\gamma_2(t),\gamma_3(t))$ for all $t$: it leaves $\gamma(t)$ equal to itself since it is already a norm $1$ vector, and also $\gamma_2(0)$ and $\gamma_3(0)$ since they already form an orthonormal base with $\gamma(0)$. It smoothly changes $\gamma_2(t)$ and $\gamma_3(t)$ in orthonormal vectors: we still denote $\Gamma=(\gamma,\gamma_2,\gamma_3):(-\varepsilon,\varepsilon)\to O(3)$ the modified path.

Finally: note $\Gamma'(0)=E'\in T_AO(3)$. Then $F\circ \Gamma=\pi_1|_{O(3)}\circ(\gamma,\gamma_2,\gamma_3)=\gamma$, and so $$dF_A(E')=dF_A(\Gamma'(0))=(F\circ\Gamma)'(0)=\gamma'(0)=E,$$ so $F$ is a submersion.


Hint Fix $A \in O(n)$, and notice that we can write $F$ as (the restriction to $O(3)$ of) the linear map $A \mapsto e_1 A$, where $(e_1, e_2, e_3)$ is the standard orthonormal basis on $\Bbb R^3$ and where we view the $e_i$ as row vectors. By linearity, we can identify $dF_A$ with (the restriction to $T_A O(n)$ of) $B \mapsto e_1 B$.

Now, since $A$ is orthogonal, $$T_{F(A)} S^2 = \{F(A)\}^{\perp} = \{e_1 A\}^{\perp} = \operatorname{span}\{e_2 A , e_3 A\} ,$$ so it suffices to find elements in $T_A O(n)$ that $dF_A$ maps to $e_2 A, e_3 A$. Since $T_I O(n)$ consists of skew-symmetric matrices, right-invariance gives that $$T_A O(n) = (dR_A)_I \cdot T_I O(n) = \{X A : X^{\top} = -X\} .$$ (Here, $R_A$ is the right multiplication map $C \mapsto CA$.)

Additional hint Thus, we just need to find skew-symmetric matrices $X_2, X_3$ whose first rows are respectively $e_2, e_3$, so that $$dF_A(X_i A) = e_1 X_i A = e_i A, \qquad i = 2, 3 .$$ We can just take $$X_2 = \pmatrix{\cdot&1&\cdot\\-1&\cdot&\cdot\\ \cdot&\cdot&\cdot}, \qquad X_3 = \pmatrix{\cdot&\cdot&1\\ \cdot&\cdot&\cdot\\ -1&\cdot&\cdot} .$$ We can also see immediately that $$\ker dF_A = \operatorname{span}\left\{\pmatrix{\cdot&\cdot&\cdot\\ \cdot&\cdot&1\\ \cdot&-1&\cdot} A\right\} .$$


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