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Is there any general rule to find the number of linearly independent equations such that $$L_i(T_{\mu\nu},\partial_\eta T_{\mu\nu},\partial_\omega\partial_\eta T_{\mu\nu},...)=0$$ where $L_i$ is a multi-linear function.

Here is a simple example: Given equations $$\partial_i b_{jk}+\partial_j b_{ki}+\partial_k b_{ij}=0$$ where $i,j,k\in\{1,2,...,n\}$ and $B=\{b_{ij}\}$ is anti-symmetric matrix. How do I know how many independent equations there are? I don't think it is $C(n,3)$ since I don't use the property of anti-symmetric matrix.

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In this generality, it is not at all clear what you mean by 'any general rule'. Of course, there is a general rule: Compute the rank of the appropriate matrix or linear map. However, computing that rank can easily be nontrivial, and there is no universal way to simplify this problem.

In your specific case, of the system of equations on the $\partial_kb_{ij}$, the number of independent equations is $C(n,3)$ after all, because you are asking for the rank the exterior multiplication mapping $$ V\otimes\Lambda^2(V)\to\Lambda^3(V). $$ This mapping is, of course, onto (because the map isn't zero and the right hand side is an irreducible $\mathrm{GL}(V)$-module), so the rank is the dimension of $\Lambda^3(V)$, which is $C(\dim V, 3)$.

As an example, though, of a case in which you don't necessarily know the correct answer right away, consider the case of the Bianchi identities $$ R_{ijkl}+R_{iklj}+R_{iljk} = 0 $$ where $R_{ijkl}=-R_{jikl}=-R_{ijlk}$. It is less obvious how many independent equations this is because you may not know the range of the corresponding map $$ \Lambda^2(V)\otimes\Lambda^2(V)\to V\otimes\Lambda^3(V) $$ because the right hand side isn't irreducible as a $\mathrm{GL}(V)$-module. (However, it turns out that it is indeed onto in this case.)

As this discussion suggests, sometimes representation theory helps, but it's not a panacea.

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  • $\begingroup$ 1. It's pretty interesting that you directly regard anti-symmetry as exterior space. But how do you handle symmetry case? 2. I think $C(n,3)$ is still suspected. Think about Poincare lemma, the equations I've given is necessary and sufficient condition to solubility of pdes $\partial_i a_j-\partial_j a_i=b_{ij}$. Doesn't this indicate only $C(n,2)$ equations independently? An equivalent transformation of question shouldn't vary on rank. $\endgroup$
    – Shuchang
    Commented Sep 20, 2013 at 9:49
  • $\begingroup$ @ShuchangZhang: 1. Symmetry is handled the same way as antisymmetry. If one had assumed $b_{ij}=b_{ji}$, then the map would have been $V\otimes S^2(V)\to S^3(V)$, which is also surjective, so this is $C(n{+}2,3)$ equations on the $C(n{+}1,2)$ unknowns. 2. You are confusing (algebraic, linear) independence of the differential equations with the generality of the space of solutions; there is no simple rank/dimension relation for PDE, as there is in case of purely algebraic linear equations. Instead, for PDE, one needs É. Cartan's subtle notion of the 'generality' of the space of solutions. $\endgroup$
    – Robert Bryant
    Commented Sep 20, 2013 at 11:52
  • $\begingroup$ @ShuchangZhang: (cont) You probably have not thought about what 'independent' really means in this context. The equations you gave in your example are independent in the sense that, if you omit any of them, then the space of solutions of the remaining equations is properly larger than the space of solutions of the full set. On the other hand, the equations certainly do satisfy differential relations: Set $E_{ijk}=\partial_ib_{jk}+\partial_jb_{ki}+\partial_kb_{ij}$, then $$\partial_iE_{jkl}+\partial_jE_{kli}+\partial_kE_{lij}+\partial_lE_{ijk}=0;$$ they aren't differentially independent. $\endgroup$
    – Robert Bryant
    Commented Sep 20, 2013 at 12:02
  • $\begingroup$ 1. I'm not quite understand the range part. For the domain it means $V$ as $e_i\partial_i$ and $S^2(V)$ means $b_{ij}e_i e_j$, what and why is $S^3(V)$? Something like $c_{ijk}e_i e_j e_k$ where $e_i e_j=e_j e_i$ for every index pair? And how to know it is a surjection? $\endgroup$
    – Shuchang
    Commented Sep 20, 2013 at 12:35
  • $\begingroup$ @ShuchangZhang: Basically, that's right. It's a surjection because the mapping $V\otimes S^2(V)\to S^3(V)$ is $\mathrm{GL}(V)$-equivariant, so the image has to be a $\mathrm{GL}(V)$-invariant subspace of $S^3(V)$. However, $S^3(V)$ is $\mathrm{GL}(V)$-irreducible, i.e., the only such subspaces are $\{0\}$ and all of $S^3(V)$, and the map clearly isn't zero, so it must be onto. $\endgroup$
    – Robert Bryant
    Commented Sep 20, 2013 at 12:40

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