How to count degrees of freedom in a symmetric $N \times N$ matrix? I am reading Wayne Hu's short lecture on cosmology mathematical infrastructure (https://arxiv.org/abs/astro-ph/0402060), and have several questions.
Some background for us lazy people that don't want to open the link:
Wayne basically shows us how to quickly count the number of degrees of freedom, using a couple of methods:

*

*A 4x4 symmetric matrix (i.e. the space-time metric, and stress-energy tensor in this case) has $\frac{4\cdot 5}{2}=10$ degrees of freedom.

*These degrees of freedom can be decomposed into Scalar, Vector and Tensor (SVT) components.

*Wayne decomposes said DOF thus:
$$g^{00}=-a^{-2}(1-2A);\;A\text{ is a scalar}\\
  g^{0i}=-a^{-2}B^{i};\; B^i \text{ a vector shift} \\
  g^{ij}=a^{-2}\left(\gamma^{ij}-2H_{L}\gamma^{ij}-2H_{T}^{ij}\right);\;$$
where $H_L$ is a scalar perturbation to the spatial curvature and $H_T^{ij}$ is a trace-free distortion to spatial metric.

Similarly, the $T^{\mu}_{\nu}$ breakdown is:
$$ T^0_0 = -\rho -\delta\rho\\
   T^0_i=(\rho+p)(v_i+B_i)\\
   T^i_0=-(\rho+p)v^i\\
   T^i_j=(p+\delta p)\delta^i_j +p\Pi^i_j$$
So far so good. However, on page 7 near the bottom, he counts the DOF thusly:
20 variables, -17 (Homogenous and Isotropic), -2 Einstein equations, -1 conservation equations, +1 Bianchi identity = 1 Degree of freedom.

*

*The (slightly stupid) first question is: Isotropy means there are no vectors "alive", Homogeneity means no distortions - i.e. no tensor "alive". That leaves us with 4(!!) scalars. I am GUESSING Homogeneity also ties in $A$ to $H_L$ but I guess I'm searching for someone to verify this.


*The second question is more involved (I think). Suppose I am dealing with a $5\times5$ matrix.
The "SVT" decomp gives us: (1) scalar $\times1$, (3) vector $\times1$, (5) Tensor of degree 2 $\times1$, (7) Tensor of degree 3 $\times1$ and (9) Tensor of degree 4 $\times1$ = 1+3+5+7+9=25.
However, enforcing symmetry (after all we are dealing with a metric) gives us $\frac{5\cdot6}{2}=15$ DOF. In the $4\times4$ case out of the scalar $\times1$, vector $\times1$, tensor $\times1$ and tensor 3-degree $\times1$, the symmetry "turns off" all of the 3rd degree tensor contributions except the scalar degree of freedom. And given that the $4\times4$ symmetric case is embedded in the $5\times5$ symmetric one we have 5 more degrees of freedom to account for. Now I am guessing the scalar of the 4th degree tensor is activated, and I can guess the remaining 4 degrees of freedom activate the 2 vectors and two tensors in the 3rd degree tensor.
Can anyone provide a methodical way of accounting for the "turned on/off" degrees of freedom in the general $N\times N$ symmetric matrix?
 A: OK,
So I figured out my own misconception.
Since the $N\times N$ matrix STILL has only two indexes, it can't embed any tensors over a 2nd degree.
Thus my notion of a 3rd degree tensor is simply not correct.
The right way to think about it is that in a $N\times N$ space-time, vector degrees of freedom have $N-1$ variables. They have $N$ entries, but the transformation rule that determines them to be vectors enforce one variable to be dependent on the other $N-1$.
By the same token, a general tensor now has a $N-1 \times N-1$. However, enforcing symmetry for the tensor means $\frac{(N-1)N}{2}$ variables, and the traceless condition removes an additional degree.
Let's test this on the $4\times 4$ case:
We have (1) scalar, (3) vector and ($3\cdot 4 /2 -1 = 5$) tensor.
The additional scalar comes from the trace of tensor that we took away previously.
All in all: $1 + 3 +5 +1 =10$
Let's look at a 5-dimensional square symmetric matrix:
We have (1) scalar, (4) vector, and ($4*5/2 -1=9$) tensor.
Again we add another scalar to get: $(1+4+9+1=15)$.
In the general case:
(1) scalar, $(N-1)$ vector, $(N(N-1)/2 -1)$ tensor + (1) scalar.
i.e. $1+N-1 +\frac{N^2 -N}{2} -1  +1=\frac{N^2}{2} +\frac{N}{2} = \frac{N(N+1)}{2} $
What threw me off was the Laplacian eigenfunctions of the 3-dimensional space.
The Laplacian eigenfunctions of a 4-dimensional space OBVIOUSLY look different.
But they still have the same decomposition into:
$$Q_0^{0}=\text{scalar}\\
Q_1^{0}=\text{scalar}\;, Q_1^{\pm 1}=\text{vectors}\\
$Q_2^{0}=\text{scalar}\;,Q_2^{\pm 1}=\text{vectors}\; , Q_2^{\pm 2}=\text{tensors}$$.
Anyway, the conclusion is that SVT decomp works for arbitrary $N\geq 4$ dimensions.
