# Dimension of the space of symmetric traceless matrices

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?

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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

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.

$^1$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.