# Dimension of SO(n) and its generators

1. Nov 18, 2015

### spaghetti3451

The generators of $SO(n)$ are pure imaginary antisymmetric $n \times n$ matrices.

How can this fact be used to show that the dimension of $SO(n)$ is $\frac{n(n-1)}{2}$?

I know that an antisymmetric matrix has $\frac{n(n-1)}{2}$ degrees of freedom, but I can't take this idea any further in the demonstration of the proof.

Thoughts?

2. Nov 18, 2015

### Staff: Mentor

Consider its tangent space, the Lie Algebra $so(n)$, or the short exact sequence $SO(n) → SO(n-1) → S^{n-1}$ and prove it by induction.

3. Nov 18, 2015

### spaghetti3451

Well, I know that the Lie algebra $so(n)$ is the space of antisymmetric matrices, so that the basis vectors of the algebra are given by the $\frac{n(n-1)}{2}$ degrees of freedom of an $n \times n$ antisymmetric matrix. Therefore, the space $so(n)$ of the antisymmetric matrices is spanned by $\frac{n(n-1)}{2}$ generators. Therefore, the dimension of $so(n)$ is $\frac{n(n-1)}{2}$. Therefore, the dimension of $SO(n)$ is also $\frac{n(n-1)}{2}$.

But then, we considered the defining representation of the $so(n)$ algebra, i.e., the representation in terms of $n \times n$ matrices. Does that mean that if we consider representations of other dimensions, they are not necessarily all antisymmetric and hence do not lead to the required number of degrees of freedom?

4. Nov 18, 2015

### Staff: Mentor

No. Since orthogonality means the transposed matrix is as well the inverse matrix, transposing respects group homomorphisms and therefore any representation of the special orthogonal group is orthogonal of determinate 1 again. One can trivially build

$SO(n) → GL(n) → GL(n+1) → ...$

but that won't change dimensions. Moreover the dimension of $SO(n)$ is defined by the dimension of its underlying manifold which isn't affected by any representations of $SO(n)$.

5. Nov 18, 2015

### spaghetti3451

So, what you are saying is that representations of any dimension are possible for $SO(n)$ and that they are all not necessarily antisymmetric.

But, since the number of dimensions of $SO(n)$ is a property of the underlying manifold of $SO(n)$, the number of dimensions of $SO(n)$ remains is the same for all representations.

Is that it?

6. Nov 18, 2015

### Staff: Mentor

Yes and ("... and that they are all not necessarily antisymmetric.") NO!

For any representation φ you can always build a representation

$$\begin{pmatrix} 1 & 0 & ... & 0\\ 0 & 1 & ... & 0\\ ...& ...& ... & ...\\ 0 & 0 & 0 & φ \end{pmatrix}$$

However you should look up your definition of representation. Strictly speaking its an analytic group homomorphism $φ : SO(n) → GL(V)$. But sometimes it's required to be in $GL(V)_ℝ$ the real Lie group underlying the complex Lie group $GL(V)$.

Nevertheless if you read my post again, I've proved that still $φ(X) φ(X)^τ = 1$ holds!

7. Nov 18, 2015

### spaghetti3451

When I asked if representations of any dimension are possible and if they are not necessarily antisymmetric, I was referring to finite-dimensional irreducible representations actually!

8. Nov 18, 2015

### Staff: Mentor

What has irreducibility to do with it?

(https://en.wikipedia.org/wiki/Spin_representation)

Last edited: Nov 18, 2015