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If $M$ is an $n \times n$ matrix, $|\det(M)|$ is the volume of the $n$-dimensional parallelepiped spanned by the column vectors of $M$.


          Parallelepiped
          (Image from Wikipedia's Determinant article.)
In the 19th-century, Cayley defined the hyperdeterminant of a hypermatrix $H$. (A hypermatrix can be viewed as a representation of a tensor.)

My question is:

Q. Does the hyperdeterminant have a geometric interpretation somehow analogous to the parallelepiped-volume interpretation of the determinant?

Perhaps an answer resides in the book by Gelfand, Kapranov, and Zelevinsky entitled Discriminants, Resultants and Multidimensional Determinants (Birkhäuser, Boston, 1994; MAA link), which I have yet to examine.

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    $\begingroup$ Umm, Wikipedia has an entry (which on my reading says no to your question), and there have been some ArXiv articles computing hyperdeterminants for small cases which might also help with your question. Perhaps you mean to ask something else? Gerhard "Question Not Ready For Primetime?" Paseman, 2015.10.07 $\endgroup$ Oct 7, 2015 at 20:47
  • $\begingroup$ @GerhardPaseman: A definitive No answer would be useful (and a bit surprising [to me]). $\endgroup$ Oct 7, 2015 at 20:48
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    $\begingroup$ From the first paragraph on the Wikipedia article for hyperdeterminant: "Many other properties of determinants generalize in some way to hyperdeterminants, but unlike a determinant, the hyperdeterminant does not have a simple geometric interpretation in terms of volumes." I read that as a "No", but you may be looking for something more. Gerhard "It Is Wikipedia, After All" Paseman, 2015.10.07 $\endgroup$ Oct 7, 2015 at 20:53
  • $\begingroup$ @GerhardPaseman: Thanks for clarifying. I don't take that Wikipedia-statement as definitive, because it cannot be literally a volume, but rather a quantity akin (in some unknown-to-me sense) to a volume. $\endgroup$ Oct 7, 2015 at 21:20

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At least, in the classical case treated by Cayley, the $2{\times}2{\times}2$ hyperdeterminant, there is an interpretation in terms of volumes generalizing the classical determinant case. It goes like this:

First, recall that, when two vector spaces $U$ and $V$ have the same dimension, say, $r$, there is a polynomial map of degree $r$ $$ \mathbf{det}: U\otimes V \to \Lambda^r(U)\otimes\Lambda^r(V) $$ defined, relative to any basis $u_i$ of $U$ and $v_j$ of $V$, by $$ \mathbf{det}( a^{ij} u_i\otimes v_j) = \det(a^{ij}) \,(u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes (v_1\wedge v_2\wedge\cdots\wedge v_r) $$ where $\det$ is the usual determinant of a matrix. It is easy to see that this definition of $\mathbf{det}$ is independent of the choice of basis $u_i$ of $U$ and $v_\rho$ of $V$. Note that, since $\Lambda^r(U)$ and $\Lambda^r(V)$ are the top exterior powers of $U$ and $V$, their elements represent 'volume elements' (sometimes called 'mass elements') in $U$ and $V$ respectively, so a tensor $\alpha\in U\otimes V$ gives rise, via $\mathbf{det}(\alpha)$ a way to 'multiply' volume elements in the two vector spaces.

(Of course, in the usual linear algebra interpretation, we consider $\alpha\in U^*\otimes V$ as a linear map from $U$ to $V$, and then $$ \mathbf{det}(\alpha)\in \Lambda^r(U^*)\otimes \Lambda^r(V)\simeq \Lambda^r(V)\otimes \Lambda^r(U)^*\simeq\ "\Lambda^r(V)/ \Lambda^r(U)" $$ can be thought of as a ratio of volume forms on the two vector spaces. Even more specially, we can take $U=V^*$, and then, because $\Lambda^r(V)\otimes \Lambda^r(V)^*$ is canonically isomorphic to $\mathbb{R}$, $\mathbf{det}(\alpha)$ becomes just a scalar.)

Now, when you have three vector spaces $U$, $V$, and $W$ of the same dimension $r$, you can expand a tensor $\alpha\in U\otimes V\otimes W$ in terms of a basis $u_i$ of $U$, $v_j$ of $V$, and $w_k$ of $W$ as $$ \alpha = a^{ijk}\,u_i\otimes v_j\otimes w_k $$ and you can just regard the $w_k$ as 'indeterminates' and define $$ \mathbf{det}_{UV}(\alpha) = \det(a^{ijk}w_k)\otimes (u_1\wedge u_2\wedge\cdots\wedge u_r)\otimes(v_1\wedge v_2\wedge\cdots\wedge v_r), $$ where, now, $$ \mathbf{det}_{UV}(\alpha) \in S^r(W)\otimes \Lambda^r(U) \otimes \Lambda^r(V). $$ To go further, you need to look at particular values of $r$. What Cayley did was consider the fact that, for quadratic forms in $r$ variables, there is a well-defined discriminant mapping $\mathrm{discr}_W:S^2(W)\to S^2(\Lambda^r(W))$ that is a homogeneous polynomial of degree $r$. It is defined by $$ \mathrm{discr}_W(q^{ij}w_iw_j) = \det(q^{ij})\, (w_1\wedge w_2\wedge\cdots\wedge w_r)^2. $$

In particular, when $r=2$, we can compose to get an element $$ \mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha)) \in S^2(\Lambda^2(U))\otimes S^2(\Lambda^2(V)) \otimes S^2(\Lambda^2(W)), $$ which is a polynomial of degree $4$ in the coefficients of $\alpha$. This is (the negative of) Cayley's hyperdeterminant in the $2{\times}2{\times}2$ case. Note that it is, indeed, expressed in terms of volume forms on the three vector spaces $U$, $V$, and $W$. It's just that it is now the product of squares of volume forms. By the way, it is not hard to show that, if we had, instead, computed $\mathbf{hdet}_{VWU}(\alpha)$, we would have got the same result.

I think that, at least in the $r=2$ case, this is probably the best interpretation of the hyperdeterminant in terms of volumes.

In the case when $(\dim U, \dim V, \dim W) = (2,2,s)$ where $s > 2$, this gives an expression $$ \mathbf{hdet}_{UVW}(\alpha) = \mathrm{discr}_W(\mathbf{det}_{UV}(\alpha)) \in S^s(\Lambda^2(U))\otimes S^s(\Lambda^2(V)) \otimes S^2(\Lambda^2(W)) $$ of degree $2s$, but, of course, this vanishes identically when $s > 4$.

When you go to higher values of $r$, it's not so clear. For example, when $r=3$ (again, with all vector spaces of the same dimension $r$), there are the two Aronhold relative invariants: $Q^4: S^3(W)\to S^4(\Lambda^3W)$ (of degree $4$) and $Q^6:S^3(W)\to S^6(\Lambda^3W)$ (of degree $6$), and so we can define two expressions $$ \mathbf{Q}^4_{UVW}(\alpha) = Q^4_W(\mathbf{det}_{UV}(\alpha)) \in S^4(\Lambda^3(U))\otimes S^4(\Lambda^3(V)) \otimes S^4(\Lambda^3(W)) $$ of degree $12$ and $$ \mathbf{Q}^6_{UVW}(\alpha) = Q^6_W(\mathbf{det}_{UV}(\alpha)) \in S^6(\Lambda^3(U))\otimes S^6(\Lambda^3(V)) \otimes S^6(\Lambda^3(W)) $$ of degree $18$. Thus, these clearly relate powers of volume elements of the underlying vector spaces. According to the Wikipedia page, though, the hyperdeterminant of $\alpha$ in this case must have degree $36$; presumably it is a polynomial in the above two invariants.

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  • $\begingroup$ Perhaps (for Joseph's definition of "somehow analogous") this could be a "Yes" answer. For those of us still struggling with the concepts, could you confirm/refute the assertion that "the classical case considered by Cayley" as you put it is the "second hyperdeterminant" as so-named in the Wikipedia article? Gerhard "Still Learning Geometry Of Hypermatrices" Paseman, 2015.10.08 $\endgroup$ Oct 8, 2015 at 15:54
  • $\begingroup$ @GerhardPaseman: Yes, the classical case considered by Cayley that I was referring to is, indeed, the second hyperdeterminant. However, I think that, in the $2{\times}2{\times}2$ case that I was discussing at that point, Cayley's first hyperdeterminant agrees with the second hyperdeterminant. $\endgroup$ Oct 8, 2015 at 16:12
  • $\begingroup$ Thank you, Robert, for taking the time to answer in such detail and with such care. The correspondence to multiplying volume forms (or their squares) when $r=2$ is instructive and pleasing. $\endgroup$ Oct 8, 2015 at 16:50
  • $\begingroup$ One might add that the hyperdeterminant of format $(2,2,3)$ is treated in Example 3.8 of Chapter 14 in GKZ. In particular they say that it is the Chow form for the Segre subvariety $P^1\times P^1\subset P^3$. $\endgroup$ Mar 13, 2022 at 10:42
  • $\begingroup$ Robert's answer is great - I just want to add one comment, in an act of shameless self-promotion, that the degree 36 hyperdeterminant in the case of $3 \times 3\times 3$ tensors is expressed as a polynomial in the lower degree invariants, see Theorem 3, arxiv.org/pdf/1310.3257.pdf $\endgroup$ Jul 21, 2022 at 18:22
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This might help, although I can only visualize it in $3$D. In your opening sentence you write:

spanned by the column vectors of M.

The $n\times n\times n$ hyper-matrix (assuming it is non-degenerate) has columns that are linearly independent in $\mathbb{R^3}$. If we place these in $\mathbb{R^3}$ and join them up again (similar as to a parallelgram), we obtain a parallelapiped.

The hyper-determinant therefore represents the hyper-volume of the associated parallelapipeds structure created from the hyper-matrix.

EDIT

Or, consider a parallelapiped with a basis a,b,c. The vertices are formed by the linear combinations a+b,a+c,b+c,a+b+c. With a $3\times3\times3$ hypermatrix, read top to bottom, we have a collection of $9$ vectors, and the vertices of the resulting shapes are the linear combinations of these vectors.

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    $\begingroup$ I'm sorry: what are you saying? I have trouble coming up with a notion of column that would embed in 3-space. Even if you embed it in n-space, I don't see that the n^2 many columns are linearly independent for n > 1. Gerhard "Please, What Is A Column?" Paseman, 2015.10.07 $\endgroup$ Oct 7, 2015 at 21:09
  • $\begingroup$ you take them slice at a time. i can't see it beyond 3d as i said. perhaps you have to take all permutations or something. $\endgroup$
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  • $\begingroup$ I'm afraid slice is also not useful to me. For n=3, let there be a hypermatrix of 3x3x3=27 entries. I pick 3 of them at a time to embed in three space. I partition the hypermatrix entries into 9 such groups which I call "vectors". What do I do with these 9 vectors in 3-space that allows me to talk about linear independence? Take the vectors 3 at a time? Gerhard "Really, What Is A Slice?" Paseman, 2015.10.07 $\endgroup$ Oct 7, 2015 at 21:15
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    $\begingroup$ In related questions on MathOverflow, including one Joseph asked some years ago, hypermatrices are (my interpretation) notational devices for tensors, and the formula for multiplication is derived and understood by students of hypermatrices, and bears some relation to composition of transformations. I think your statement about its having columns that are linearly independent in R^3 is wrong (I have a 0-1 example in mind), but I can't be sure since we haven't agreed on what it is you mean. Gerhard "Again, What Is A Column?" Paseman, 2015.10.07 $\endgroup$ Oct 7, 2015 at 21:27
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    $\begingroup$ @GerhardPaseman: Good memory!: "Why are matrices ubiquitous but hypermatrices rare?." ~5 yrs ago! $\endgroup$ Oct 7, 2015 at 21:42

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