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The area is (half) the product of a side by the height, so the question of why we get square roots boils down to why the height is a square root. This can be tracked down by applying Pythagoras to both right triangles made by the height.
Are $C$ and $X$ simple bivectors (i.e. have minimum rank, which is $2$ for an antisymmetric matrix), or general ones, in which case seeing them as bivectors does not restrict the problem further?
Digging a bit in the literature it seems people care mostly about the low-rank case, namely embedding into a Euclidean space of a fixed dimension and asking about complexity as the graph size grows. This leads to an NP-hard problem. Did you find interesting references that do not impose a rank condition?
Of course, the probably-unknown complexity of your problem in the bit model (namely counting bits used by coordinates) does not tell us anything about the complexity in the real model (namely ignoring issues of numerics). This discussion of NP should not be confused with the known fact (irrelevant to the present discussion) that the same question with a fixed bound on the matrix rank is NP hard.
This problem goes by the name "Positive Semi Definite Matrix Completion", see for instance the 1997 review by Monique Laurent. It is related to the question of whether a graph with given edge lengths can be embedded into Euclidean space, which is amenable to semi-definite programming (SDP). However, it seems people don't know bounds on the "number of digits" of a solution to the SDP, so that the problem may even fail to be in NP. Do you care about such numerical artifacts, or only about the combinatorics problem itself?
The condition "$M$ is PSD in its specified entries" is still missing from this answer: a non-PSD matrix with all entries specified gives a trivial counterexample of what is stated here.
Thanks for the detailed answer, I had mistaken impressions about constructive mathematics. Just a point of detail: the proof that $S_2$ is large depends on bounding below a sum of terms, which are still non-negative in your variant, and bounding below the first term $n=0$, namely $\frac{1+\cos\pi x_m}{1+x_m} > \frac{1-\sin\pi q}{3/2+q}$, with a slightly worse numerator than you wrote. This doesn't change the final conclusion, but the choice of $q$ is more annoying to state.
E.g., for $D_n$, consider the representation with $\mu=\varpi_{n-1}+\varpi_n$ namely the representation $R=\Lambda^{n-1}V$ where $V$ is the defining representation. Then $S^2 R$ and $\Lambda^2 R$ both contain $\text{adj}(G)$, contrarily to your conjecture, in contrast to the case of all $R=\Lambda^kV$ for $k\leq n-2$, for which your conjecture is correct. I still need to work out a bunch of details and see if there are natural generalizations, and I will then write an answer to my own question in a month or so.
Thank you very much! I found the answer and will write it up ASAP. I explicitly determined the $\text{adj}(G)$ inside $R\otimes\overline R$, labeled by non-zero entries $\mu_i$ of the highest weight $\mu=\sum_i\mu_i\varpi_i$ in the basis of fundamental weights $\varpi_i$. If $\varpi_i\leftrightarrow\varpi_j$ under conjugation then any (real) representation with $\mu_i=\mu_j>0$ will have a pair of $\text{adj}(G)$ in $S^2R$ and $\Lambda^2R$. If $\varpi_i$ is conjugation-invariant then the corresponding $\text{adj}(G)$ sits in $\Lambda^2R$ or $S^2R$ if $R$ is real/quaternionic.
