# Show that a specific map is a submersion of $O(3)$ in $S^2$

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 $$\ker dF_A = \operatorname{span}\left\{\pmatrix{\cdot&\cdot&\cdot\\ \cdot&\cdot&1\\ \cdot&-1&\cdot} A\right\} .$$