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Background

I have a set $S$ (that is possibly infinite) and a correspondence between functions $c:S^3\to\mathbb{R}$ (I will write $c(i,j,k)$ as $c_{ijk}$) and matrices $M$ with rows indexed by $(i,j,k)\in S^3$ and columns indexed by $(m,n)\in S^2$ defined as follows:

$$M_{(i,j,k),(m,n)}=\delta_{im}c_{njk}+\delta_{jm}c_{nik}-\delta_{kn}c_{ijm}$$

where $\delta_{ij}$ is the kronecker delta function.

Question

I want to determine some necessary and sufficient conditions on $c$ so that the above matrix has trivial null space.

Any references, ideas, techniques, etc. would be appreciated. Of course, I'm not looking for a complete solution to the problem, I just don't know what sort of methods might even be used to attack this (despite it appearing at first to be a simple linear algebra problem).

Progress

For finite $S$ with cardinality $N$, this is a $N^3\times N^2$ size matrix, so it should be true that most choices of $c$ result in a matrix with trivial null space. However, I'm still not sure what "most" would even mean in this context.

I've also proven the result for $c_{ijk}=\delta_{ijk}$ (that is $c_{iii}=1$ and equals zero otherwise), but otherwise have no further results.

Motivation

I am investigating $\mathbb{R}$-algebras $\mathcal{A}$ equipped with a "derivation" operator $\mathcal{D}:\mathcal{A}\to\mathcal{A}$, which is simply a linear operator satisfying the product rule. That is, $\mathcal{D}(a\cdot b)=\mathcal{D}(a)\cdot b+a\cdot\mathcal{D}(b)$ for any $a,b\in\mathcal{A}$.

We can first consider a basis $\{b_i\}_{i\in I}$. Then let $c_{ijk}$ be the coefficient of $b_k$ when $b_i\cdot b_j$ is written using the basis elements. Next, since every derivation is determined by its action on basis elements, we can let $\mathcal{D}(b_i)=\sum_j a_{ij}b_j$ (where this sum is nonzero on finitely many $j$ and thus is well-defined).

Next, we can note that a derivation is simply an element $(a_{ij})$ of the kernel of the above matrix. My conjecture is that whenever $\mathcal{A}$ is finite, there are no nonzero derivations. In the infinite case, I'm just completely lost, so any direction of attack would be appreciated.

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  • $\begingroup$ I'm not an expert on what might happen in the infinite dimensional case, but the finite dimensional case is really more appropriate for MSE at least. Also, matrices without kernel are open and dense in the collection of all $m \times n$ matrices, so this handles your question of what "most" should mean. Finally, I believe you have a typo in your question: there is no defintion of $l$ anywhere and yet it is a variable in your first big formula. $\endgroup$
    – A. Thomas Yerger
    Commented Jun 21, 2021 at 0:46
  • $\begingroup$ Thanks for spotting the typo. Seems like “open and dense” should be the appropriate notion for “most” here yes. I don’t think we immediately get even this, though, and certainly characterizing the “good” c’s even in the finite dimensional case doesn’t strike me as a problem that is likely to be elementary enough for MSE. Perhaps I am wrong, though, and you see an easy line of attack to get some sort of useful result out of this. Btw, nice to see the >implying founder here (: $\endgroup$
    – exfret
    Commented Jun 21, 2021 at 1:03
  • $\begingroup$ In any case, the fact that S can be infinite should be enough to promote this to “research-level”, at least. $\endgroup$
    – exfret
    Commented Jun 21, 2021 at 1:05
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    $\begingroup$ The finite dimensional case is "trivial" in the sense that for a given $c$ it is easy to hand to a computer and get the answer. What sort of a characterization are you looking for? $\endgroup$
    – A. Thomas Yerger
    Commented Jun 21, 2021 at 2:48
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    $\begingroup$ There are situations where you can think of such operators as "infinite matrices," I think one example is when you're dealing with e.g. $\ell^2(\mathbb{N})$? I think the point in such examples is that really you're just working with a separable Hilbert space, so you can get a countable basis and orthonormalize it. In fact I think once you've got this setup, the same proof as in finite dimensions will allow you to see that the operators with kernel 0 are open and dense. But I should not embarrass myself anymore and let someone who knows shed light on the functional analysis. $\endgroup$
    – A. Thomas Yerger
    Commented Jun 21, 2021 at 20:07

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