show that the isotropy group of r in SO(3) is isomorphic to SO(2)

**oblixps** · November 14th 2011, 11:46 PM

so what i'm trying to show is that the set of all 3x3 matrices with determinant = 1, such that for a fixed vector r, Ar = r is isomorphic to the set of all 2x2 matrices with determinant = 1.

since r is an eigenvector with eigenvalue 1 of the 3x3 matrix, the other 2 eigenvalues of the 3x3 matrix must multiply to 1 and i was thinking about how to map each 3x3 matrix with eigenvalues 1, $\lambda_1$ , $\lambda_2$ to a 2x2 matrix with eigenvalues $\lambda_1$ and $\lambda_2$ . however i am having trouble coming up with the explicit mapping. I tried taking a 3x3 matrix with those specified eigenvalues and thought about how to "trim the edges" (for example, remove the first column and first row) to get a 2x2 matrix with the same eigenvalues as the eigenvalues besides the 1 of the 3x3 matrix. but this process may "mess up" the determinant for some 3x3 matrices and end up creating a 2x2 matrix with different eigenvalues then the one i wanted. furthermore i don't even think that this is well defined.

can someone give me some hints in the right direction on how to define such a function from one set to the other? thanks.

**NonCommAlg** · November 15th 2011, 05:44 AM

Originally Posted by oblixps

so what i'm trying to show is that the set of all 3x3 matrices with determinant = 1, such that for a fixed vector r, Ar = r is isomorphic to the set of all 2x2 matrices with determinant = 1.

since r is an eigenvector with eigenvalue 1 of the 3x3 matrix, the other 2 eigenvalues of the 3x3 matrix must multiply to 1 and i was thinking about how to map each 3x3 matrix with eigenvalues 1, $\lambda_1$ , $\lambda_2$ to a 2x2 matrix with eigenvalues $\lambda_1$ and $\lambda_2$ . however i am having trouble coming up with the explicit mapping. I tried taking a 3x3 matrix with those specified eigenvalues and thought about how to "trim the edges" (for example, remove the first column and first row) to get a 2x2 matrix with the same eigenvalues as the eigenvalues besides the 1 of the 3x3 matrix. but this process may "mess up" the determinant for some 3x3 matrices and end up creating a 2x2 matrix with different eigenvalues then the one i wanted. furthermore i don't even think that this is well defined.

can someone give me some hints in the right direction on how to define such a function from one set to the other? thanks.

let $k$ be the base field. of course we also need $r \neq 0.$ let $G=\{A \in SO(3): \ Ar=r \}.$ since $r \neq 0,$ there exists a basis $\{r,u,v\}$ of $k^3$ and, with respect to that basis, every element of $G$ is in the form

$A=\begin{pmatrix} 1 & a & b \\ 0 & x & y \\ 0 & z & t \end{pmatrix},$

for some $a,b,x,y,z,t \in k.$ now since $A \in SO(3),$ we have $B = \begin{pmatrix} x & y \\ z & t \end{pmatrix} \in SO(2)$ and $a =b=0.$ so the group isomorphism you're looking for is the map $A \mapsto B.$

**oblixps** · November 16th 2011, 12:55 AM

it seems to me that this mapping is not one to one nor well defined. If i take the matrix A you wrote and replaced a, b with some other numbers then i would have 2 different matrices in SO(3) mapping to the same matrix B in SO(2). why did you say a = b = 0?

**Deveno** · November 16th 2011, 04:44 AM

but you can't replace a and b with "other numbers". A is supposed to be orthogonal, not just a matrix with determinant 1. this means that AᵀA = I,

and a quick calculation shows that the 1,1-entry of AᵀA is $1+a^2+b^2$ , so we have to have a = b = 0.

**NonCommAlg** · November 16th 2011, 05:32 AM

Originally Posted by Deveno

but you can't replace a and b with "other numbers". A is supposed to be orthogonal, not just a matrix with determinant 1. this means that AᵀA = I,

and a quick calculation shows that the 1,1-entry of AᵀA is $1+a^2+b^2$ , so we have to have a = b = 0.

that of course wouldn't work if, for example, $k = \mathbb{C}.$ so one should argue like this:since rows 1 & 2 and 1 & 3 of $A$ are orthogonal, we have $B \begin{pmatrix} a \\ b \end{pmatrix} = 0.$ thus $a=b=0,$ because $\det B = \det A = 1 \neq 0.$

**Deveno** · November 16th 2011, 07:54 AM

yes, but i think there is another name for that structures besides SO(3), that is: SU(3), in which case one doesn't use the transpose, but rather the hermetian adjoint, and you STILL get:

$1+|a|^2 + |b|^2 \implies |a| = |b| = 0 \implies a = b = 0$

in "some" way of looking at it, the type of matrix you are using, and the inner product you are calculating (since we are talking about orthogonality, we have an inner product) amount to the same thing; that is, it makes no difference if you define orthogonal matrices by:

<Ux,Ux> = <x,x> or by $U^{\dagger}U= I$ , you get the same set of matrices.

in fact, shouldn't you actually write:

$B\begin{bmatrix}\overline{a}\\ \overline{b}\end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix} \implies \overline{a} = \overline{b} = 0 \implies a = \overline{\overline{a}} = \overline{0} = 0$ ,

and likewise for b? because xa + yb is NOT the complex inner-product of the first and second rows of A.

**oblixps** · November 16th 2011, 11:16 PM

hm when calculating $A^{T}A$ i don't seem to get the entry $1 + a^2 + b^2$ anywhere. that seems to come up in $AA^{T}$ instead. but when A is orthogonal $AA^T = A^{T}A$ right?

**Deveno** · November 16th 2011, 11:39 PM

um...err....yup.

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