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In an $\mathbb{R}$-vector space $V$, the scalar product is a paradigmatic example of a non-degenerate, symmetric, positive-definite bilinear form $\beta : V \times V \to \mathbb{R}$.

I wonder if the scalar product can be generalized to more than two vectors. I am particularly eager to know if an operation analogous to the scalar product can be defined, which is, however, based on an $m$-multilinear form $\mu : V^m \to \mathbb{R}$ with $m > 2$. If so, what could be the geometrical interpretaion of such a "multilinear scalar product"?

Edit: If possible, the properties of the ordinary scalar product should be preserved.

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  • $\begingroup$ Determinant $det:\mathbf{R}^{n}\times \mathbf{R}^{n}\times...\times \mathbf{R}^{n}\rightarrow \mathbf{R}^{n}$ is nondegenerate and multilinear, though unfortunately, it is alternating rather than symmetric. So, the analogy breaks down a bit. $\endgroup$ – Sinister Cutlass Nov 25 '15 at 14:12
  • $\begingroup$ yes I just see it $\endgroup$ – janmarqz Nov 25 '15 at 14:13
  • $\begingroup$ Well, fair enough, the OP didn't indicate whether or not he wanted to preserve all properties of the scalar product. $\endgroup$ – Sinister Cutlass Nov 25 '15 at 14:14
  • $\begingroup$ @SinisterCutlass: Ah sorry. If possible, the properties should be preserved. $\endgroup$ – Björn Friedrich Nov 25 '15 at 14:18
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    $\begingroup$ Just an inkling, but if a nice answer to this question exists, I'm fairly sure it will have something to do with the tensor product of vector spaces (using either the symmetric or antisymmetric tensor product, probably). $\endgroup$ – Omnomnomnom Nov 25 '15 at 15:01
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Let me explain supposing that $m=2$ first.

If you consider the set of all bilinear maps $V\times V\to\Bbb R$, with $V={\rm span}\{b_i\}$, one should also consider that the effect of taking a pair of basic duals $\beta^j$ such that $\beta^j({b_i})=\delta^j{}_i$.

With pair of $\beta^i,\beta^j$ one gets $$\beta^i\otimes\beta^j:V\times V\to\Bbb R$$ given by $$\beta^i\otimes\beta^j(v,w)=\beta^i(v)\beta^j(w).$$ This map is bilinear. It is used to forge out all bilinear maps into a vector space generated by $\{\beta^i\otimes\beta^j\}$, could we dubbed this space as $V^*\otimes V^*$?

So, this gives us an representation of a bilinear map $T:V\times V\to\Bbb R$ in the style $$T=T_{sr}\beta^s\otimes\beta^r$$ One can then take advantage of a certain interior product on $V$ are given.

That is, if an inner product $\bullet:V\times V\to\Bbb R$ is given one forms its metric tensor (matrix) $$G=[g_{ij}],$$ where $g_{ij}=b_i\bullet b_j$. This construction allow to calculate for any $v,w\in V$: $$v\bullet w=v^sv^rg_{sr}.$$

With this idea one define for $f,\phi$ a pair of covectors $$f\bullet\phi=f_s\phi_rg^{sr}$$ where $g^{sr}$ are entries of the matrix $G^{-1}$.

Now for a pair of contravariant tensors $T=T_{sr}\beta^s\otimes\beta^r$ and $U=U_{sr}\beta^s\otimes\beta^r$, one can guess that a construction like $$T\bullet U:=T_{ij}U_{kl}g^{ik}g^{jl}$$ is useful as an analog of a inner product but now in $V^*\otimes V^*$.

Finally for the $m$-multilinear maps $V^m\to\Bbb R$, one is compelled to, in the moment of consider two of them, say, $$\mu=\mu_{i_1...i_m}\beta^{i_1}\otimes\cdots\otimes\beta^{i_m},$$ and $$\theta=\theta_{i_1...i_m}\beta^{i_1}\otimes\cdots\otimes\beta^{i_m},$$ to take as their inner product the number $$\mu\bullet\theta=\mu_{i_1...i_m}\theta_{j_1...j_m}g^{i_1j_i}\cdots g^{i_mj_m}.$$

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