Reference request: Homogenous Polynomials over graphs? Let $V$ be a set of ordered vertices such that $|V|=n$. There are $2^{n^2}$ possible directed graphs on $V$. Let $U\subset V$ such that $|U|=k$, any such subset again has $2^{k^2}$ possible directed graphs on it. We label any graph on the subset $U$ as $n_i$, where $i$ is the $i^{th}$ graph in an enumeration of the possible $2^{k^2}$ graphs. Now given an instance of
a graph $(V,E)$, I can associate a monomial to this graph as follows:
$$M = \prod_{U\subset V \\ |U|=k} n_{i}^{\alpha_i}$$
Basically, $\alpha_i$ is the number of times the $k-sized$ subgraph labelled $n_i$ appears in the graph (upto isomorphism).
Now, if I were to sum all such monomials for all $2^{n^2}$ graphs, I will get the following polynomial:
$$P(V) = \sum_{\bf{\alpha}} C_{\alpha}\prod_{U\subset V\\|U|=k}n_{i}^{\alpha}$$
Where $\alpha = <\alpha_1 ...\alpha_{2^{k^2}}>$, is the vector of cardinality of such subgraphs.
Is there any efficient way to generate $P(V)$ ? Is there any literature on this problem ?
Notice that these polynomials are always homogeneous with degree $n \choose k$
PS: Please leave a comment if the question is not clear

My definition of subgraph on a subset $U \subset V$ of a directed graph $(V,E)$ is: $(U,E')$ such that $E'$ are all the edges in $(V,E)$, which were defined on $U$.
 A: To approximately estimate $C_\alpha$, it is convenient to view $C_\alpha/ 2^{n^2}$ as the probability that a random complete directed graph / tournament on $V$ has subgraph cardinalities $\alpha_1,\dots, \alpha_{2^{k^2}}$.
A first step is compute the moments, for a vector of exponents $e_1,\dots, e_{2^{k^2}}$, $$\sum_\alpha \frac{C_\alpha}{2^{n^2}} \prod_{i=1}^{2^{k^2}} \left(\alpha_i- \frac{ \binom{n}{k}} { 2^{k^2}}\right)^{e_i}.$$
One can think of these as certain derivatives of your polynomial. Here I subtracted the expectation $\frac{ \binom{n}{k}} { 2^{k^2}}$ of $\alpha_i$ to get the most interesting moments.
Let $1_i(U)$ be the random variable that equals $1$ if the graph restricted to $U$ is the $i$th graph and $0$ otherwise. Then $\alpha_i = \sum_U 1_i(U)$ so $$\alpha_i - \frac{ \binom{n}{k}} { 2^{k^2}} = \sum_U \left(1_i (U) - \frac{1}{2^{k^2} } \right).$$
Thus we can expand this moment out as
$$ \sum_{ \substack{U_{1,1},\dots, U_{1,e_1}, U_{2,1}, \dots, U_{2^k, e^{2^k}} \subset V \\ | U_{i,j} | = k}}  \frac{1}{ 2^{n^2}} \sum_{\alpha} \prod_{i=1}^{2^k} \prod_{j=1}^{e_i} \left( 1_i (U_{i,j}) - \frac{1}{2^{k^2}} \right).$$
Now if there is any $U_{i,j}$ which doesn't share a vertex with the other $U_{i',j'}$, the inner sum over $\alpha$ will vanish because $1_i ( U_{i,j})- \frac{1}{2^{k^2}}$ will be independent of the other variables and since its expectation is $0$ this will force the sum to cancel.
The number of remaining terms is $O_{k,e} ( n^{ (k-1/2) \sum_i e_i})$ because we have $k \sum_i e_i$ elements of the $U_{i,j}$ to choose but there must be at least $ \sum_i e_i/2$ pairs that share a vertex, so $n^{ (k-1/2) \sum_i e_i})$ independent vertices in total.
Since for each fixed tuple of $U_{i,j}$, the sum over $\alpha$ contributes at most $1$, the largest contributions will come from the sets $U_{i,j}$ where each $U_{i,j}$ shares exactly one vertex with one other $U_{i,j}$.
Using this, I think, but am not completely sure, that one can show that as $n$ goes to $\infty$, the moment divided by $ n^{ (k-1/2) \sum_i e_i}$ converges to the moments of a Gaussian random variable with a variance-covariance matrix that can be calculate by considering the correlations between $\alpha_i (U_{i,j})$ and $\alpha_{i'} (U_{i',j'})$ when $U_{i,j}$ and $U_{i',j'}$ share exactly one vertex.
If this is true, then the normalized deviations $$ \frac{\alpha_i - \frac{ \binom{n}{k}} { 2^{k^2}}}{n^{ k- \frac{1}{2}}}$$ will converge as $n$ goes to $\infty$ to a Gaussian random variable with the same mean and covariance matrix.
This should give the largest-scale behavior of this polynomial.
