Intuition on Kulkarni-Nomizu product

Question

$\renewcommand\vec[1]{{\boldsymbol #1}}$If $V$ is a vector space and $T,S \in \mathfrak{T}^0_{\;\;2}(V)$ are covariant tensors of type $(0,2)$, then we define their Kulkarni-Nomizu$\newcommand\KN{\bigcirc \kern-2.5ex\wedge \;}$ product $T \KN S \in \mathfrak{T}^0_{\;\;4}(V)$ by $$(T \KN S)(\vec{x},\vec{y},\vec{z},\vec{w}) \doteq T(\vec{x},\vec{z})S(\vec{y},\vec{w}) + T(\vec{y},\vec{w})S(\vec{x},\vec{z})-T(\vec{x},\vec{w})S(\vec{y},\vec{z}) - T(\vec{y},\vec{z})S(\vec{x},\vec{w}).$$

I understand that $T \KN S$ is a curvature like tensor, in the sense that it satisfies the following symmetries:

1. $(T \KN S)(\vec{x},\vec{y},\vec{z},\vec{w}) = -(T\KN S)(\vec{y},\vec{x},\vec{z},\vec{w}) = -(T \KN S)(\vec{x},\vec{y},\vec{w},\vec{z})$;
2. $(T \KN S)(\vec{x},\vec{y},\vec{z},\vec{w}) = (T \KN S)(\vec{z},\vec{w},\vec{x},\vec{y})$;
3. $(T \KN S)(\vec{x},\vec{y},\vec{z},\cdot)+(T \KN S)(\vec{y},\vec{z},\vec{x},\cdot)+(T \KN S)(\vec{z},\vec{x},\vec{y},\cdot)=0$, if $T$ and $S$ are symmetric.

We also have the bonus property that $T \KN S = S \KN T$.

And although this quantity appearing frequently in certain calculations is enough justificative for giving it a name and notation, this seems artificial to me so far.

I do not have any intuition whatsoever for that formula, nor can I think of a way to write it neatly as $\sum_{\sigma \in S_4}{\rm something}$. We can write $$(T\KN S)(\vec{x},\vec{y},\vec{z},\vec{w}) = \begin{vmatrix} T(\vec{x},\vec{z}) & S(\vec{y},\vec{z}) \\ T(\vec{x},\vec{w}) & S(\vec{y},\vec{w})\end{vmatrix}-\begin{vmatrix} T(\vec{y},\vec{z}) & S(\vec{x},\vec{z}) \\ T(\vec{y},\vec{w}) & S(\vec{x},\vec{w})\end{vmatrix},$$but I'm failing to interpret this either.

I'd like some insight on this definition.

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If I didn't mess up, we have the more "symmetric" expression

$$(T\KN S)(\vec{x},\vec{y},\vec{z},\vec{w}) = \begin{vmatrix} T(\vec{x},\vec{z}) & S(\vec{x}+\vec{y},\vec{z}) \\ T(\vec{x},\vec{w}) & S(\vec{x}+\vec{y},\vec{w})\end{vmatrix}+\begin{vmatrix} S(\vec{x},\vec{z}) & T(\vec{x}+\vec{y},\vec{z}) \\ S(\vec{x},\vec{w}) & T(\vec{x}+\vec{y},\vec{w})\end{vmatrix}.$$Still not happy.

Answer 1

$\newcommand\KN{\bigcirc\kern-2.5ex\wedge \;}$Your symmetries $1-2$ are equivalent to requiring that $T \KN S \in S^2(\Lambda^2 V)$; so what we're really trying to do is to take bilinear forms on $V$ (in practice these are usually built from the metric and the Ricci tensor) and turn them into bilinear forms on $\Lambda^2 V$.

Any symmetric bilinear form $T \in S^2 V$ naturally induces a symmetric bilinear form on two-forms $\Lambda^2 T \in S^2 (\Lambda^2 V)$ via the formula $$(\Lambda^2 T)(v \wedge w, x \wedge y) = T(v,x)T(w,y) - T(v,y) T(x,w).$$ If you're unconvinced about the naturality of this construction, note that it corresponds via raising an index to the endomorphism $\Lambda^2 T^\sharp : \Lambda^2 V \to \Lambda^2 V$ induced on two-forms by an endomorphism $T^\sharp : V \to V$, which may be more familiar.

Thinking of $\Lambda^2$ as a function $S^2V \to S^2(\Lambda^2 V)$, we see that it is in fact a (vector-valued) quadratic form. By polarizing this quadratic form (a natural thing to do!) we get a symmetric bilinear form on $S^2(V)$. This bilinear form is (up to a factor of $\frac12$) the Kulkarni-Nomizu product. That is, the K-N product is the unique symmetric bilinear mapping $S^2(V) \times S^2(V) \to S^2(\Lambda^2 V)$ such that $T \KN T = 2\Lambda^2 T$. (I think the factor of $2$ is an unfortunate historical accident, but it does have the convenient property of making the curvature tensor of the sphere equal to $g \KN g$.)

That's the best justification (other than the fact it pops up so often) I have for its existence - hopefully that makes it seem a bit less artificial. For real intuition I think you just have to work with it for a while - I found its role in the irreducible decomposition of curvature tensors made it click for me. I learnt this stuff from chapter 11 of this freely available book.