4
$\begingroup$

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 $\gamma(0) = A$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.