Question

Group eigenvalue thingy, is this a known concept/thing?

Answer 1

Since each column or row of a group's multiplication table can be represented by a permutation matrix, I thought I'd take this a step further and write the matrix equations for it.

Here's the multiplication table for a group, http://4.bp.blogspot.com/_glb3T9YcYXA/TGYzghatV-I/AAAAAAAAADA/gMGHL1FP2lQ/s1600/e.jpg

Now let's make the column vector

\[\vec v = \begin{pmatrix}e \\ a \\ b \\c\end{pmatrix}\]

So now we can multiply this by e, a, b, or c and it will change each of the entries like so:

\[a\vec v = \begin{pmatrix}a \\ e \\ c \\b\end{pmatrix}=\begin{pmatrix}0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \\0&0&1&0\end{pmatrix}\begin{pmatrix}e \\ a \\ b \\c\end{pmatrix}=\begin{pmatrix}0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \\0&0&1&0\end{pmatrix}\vec v\]

We can define this matrix to be \(P_a\) to represent the permutaton matrix we get from multiplying by \(a\), so we have:

\[P_a \vec v = a \vec v\]
We can see this sort of thing will be true in general that there is a coresponding permutation matrix for each group element \(g_i \in G\),
\[P_i \vec v = g_i \vec v\]
which looks an awful lot like the eigenvalue equation for a matrix with a specific eigenvector \(\vec x_i\) with eigenvalue \(\lambda_i\).
\[A \vec x_i = \lambda_i \vec x_i\]

But there's actually quite a huge difference going on now. So is this related to anything or does it have a name so I can look into it more?

Answer 2

or if you guys just wanna play around with this or ask questions about group theory I'm bored foolin' around lol

Answer 3

It kind of looks like 4 vector stuff?

Answer 4

well it is a 4 dimensional vector space, but groups can be of any size, I just picked this as an example cause groups with only 1,2, or 3 elements in them are boring. I think this specific group for my example is called the Klein 4-group.

Answer 5

Mhm yeah interesting, I think I was sort of things position time four vector haha, \[x^{\mu}, \mu = 0,1,2,3\] \[x^0=ct,~~ x^1 = x,~~ x^2 = y,~~ x^3 = z\] and we can derive lorentz transformations from this, but it probably doesn't help you in anyway xD

Answer 6

you don't have \(x^0 =-ct\) or the other three components negative? Or are you able to just throw that into the metric tensor \(g_{\mu\nu}\)?

Answer 7

idk how to derive Lorentz transformations, can you show me? :O

Answer 8

Yeah we throw that into the metric tensor haha

Answer 9

I honestly don't even know what a Lorentz transformation is but if you can show me the derivation or w/e and like explain the physical stuff I can follow. I know the math like, \(g_{\mu\nu}x^\mu x^\nu = |\vec x|^2\) (with Einstein summation notation implied) but I have only done it for like geometry sorta things like on the surface of a sphere or donut or whatever.

Answer 10

I know these are the groups guaranteed by Cayley's theorem. But the matrix stuff is interesting.

Answer 11

and by interesting I mean fun

Answer 12

Haha yeah I am definitely running into issues if I think too hard about it because then it's like I'm assuming a ring stucture or something if I do anything more than permutation matrices buuut I'm not going to think too hard just have fun for now. :P

Answer 13

it's kind of like a tensor eigenvalue equation or something. The main kind of concept I'm looking at is:

a unique group element (analogous to an eigenvalue in a sense) with one universal eigenvector which is unchanging gives a set of permutation matrices.

If the group is commutative, then

\[\vec v g_i = g_i\vec v\]

However it's not at all apparent to me yet what will happen when it's nonabelian, do we end up with this situation ever?

\[\vec v g_i = h_i \vec v\]

It seems like in a worst case scenario for a nonabelian group we end up with 2n permutation matrices with n the order of the matrix. (Ok we could clearly see there's at least the identity element meaning 2n-2 unique permutation matrices...) So kind of interesting.

Then just thinking of how many possible unique permutation matrices exist in n-dimensional space, it seems like there's a connection here between the total number of ways we can pick permutation matrices of n-dimensions to the total number of unique groups of order n which is fascinating.

Answer 14

One of lifes greatest unanswered questions

Answer 15

But we already know all abelian groups

Answer 16

That's fine there's nothing particular about this to abelian or nonabelian groups maybe it'd be helpful.

What do you mean when you say we already know all abelian groups, is that like you mean all abelian group multiplication tables are reperesented by symmetric matrices or something?

Answer 17

Interesting, just by looking at a Cayley table we can see that the things being multiplied at the top and side if included as part of the matrix are symmetric if you transpose the matrix. So this means for all groups:

\[P_i \vec v = g_i \vec v\]\[P_i^\top \vec v = \vec v g_i\]

Since abelian groups have symmetric cayley tables, \(P_i=P_i^\top\) and that is sort of just another way of expressing this fact that \(g_i\) commutes with all the elements of G contained in the vector \(\vec v\):

\[g_i \vec v = P_i \vec v = P_i^\top \vec v = \vec v g_i\]

Answer 18

cute, the cayley table matrix C is:

\[\vec v \vec v^\top =C\]