Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A set of symmetric $n \times n$ matrices have $ \frac{1}{2}n^{2} - \frac{1}{2}n$ independent elements. But how do you get to this result? I understand that a general $n \times n$ matrix would have $n^{2}$ independent elements. How exactly does the constraint of symmetry lead to the piece above? Furthermore, how would I determine the number of elements for a symmetric traceless matrix?

share|improve this question
    
This is not representation theory........ –  user38268 Oct 20 '12 at 13:31
    
it's applications lie in representation theory. for example, the Lie algebra of O(n) can be represented by real nxn antisymmetric matrices, making dim(L(O(n))) = 1/2*n*(n-1). do you have any ideas at how to determine this dimension? that would be helpful. –  johndmalcolm Oct 20 '12 at 13:47
    
The lie algebra of O(n) has the addition constraint that the trace of an element sums to zero. –  user38268 Oct 20 '12 at 13:52

1 Answer 1

I'll$^1$ take an example from $SO(N)$:

Consider $T^{ij}$ containing in total $N\times N = N^2$ indep. objects$^2$. But one can show that the symmetric, antisymmetric and the trace of $T^{ij}$ transform within themselves i.e. they don't mix with each other: $$\tag{def. trans. rule for each index}T^{ij}\rightarrow T^{'ij} = O^{il}O^{jm}T^{lm}$$ $$\tag{the symmetric part}S^{ij}\rightarrow S^{'ij} = O^{il}O^{jm}S^{lm}$$ $$\tag{the antisymmetric part}A^{ij}\rightarrow A^{'ij} = O^{il}O^{jm}A^{lm}$$ where $S^{ij} = 1/2(T^{ij}+T^{ji})$ and $A^{ij} = 1/2(T^{ij}-T^{ji})$.

Finally the trace $T:=\delta^{ij}T^{lm}$ transforms as $$T\rightarrow T' = \delta^{ij}T^{'ij} = \delta^{ij}O^{il}O^{jm}T^{lm} = (O^T)^{li}\delta^{ij}O^{jm}T^{lm} = \delta^{lm}T^{lm} = T$$ where we used $O^TO=\mathbf{1}$. Thus we can define the symmetric traceless tensor from $T^{ij}$ $$Q^{ij} := S^{ij}-\frac{1}{N}\delta^{ij}T$$ which contains $\frac{1}{2}N(N+1)-1$ objects.

In other words, one can split up the $N^2$ objects as

$$N\otimes N = [\frac{1}{2}N(N+1)-1]\oplus 1\oplus \frac{1}{2}N(N-1), $$ where the $1$ comes from the trace.

To count the objects in the symmetric and asymmetric tensors just use (draw an $N$ by $N$ matrix and count the entries including the diagonal for the symmetric one, but not for the asymmetric one):

$$\tag{for the symmetric part}\sum_{j=1}^{N} j = \frac{1}{2}N(N+1). $$ While for the asymmetric part we get $$\tag{for the antisymmetric part}\sum_{j=1}^{N} j - N= \frac{1}{2}N(N-1). $$ where the $N$ comes from the $N$ zeros on the diagonal.

See also this PSE entry.


$^1$Disclaimer: I'm not a mathematician so the below might contain errors and wrong terminology. Feel free to teach/correct me.

$^2$With the number of objects contained in a tensor we mean the dimension of the representation.

$^3$Much of this is taken from "QFT in a nutshell" by A. Zee.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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