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I was reading Schutz's book on General Relativity. In it, he says that a(n) $M \choose N$ tensor is a linear function of $M$ one-forms and $N$ vectors into the real numbers.

So does that mean the determinant of an $n \times n$ matrix is a $0 \choose n$ tensor because it is a function that maps the $n$ column vectors of the matrix to a real number (the value of the determinant)?

But then, the determinant also maps the $n$ column vectors of the matrix to the same real number (the value of the determinant).

So would the tensor representation of the determinant be different if you choose the map for the column vectors than the map for the row vectors?

mihirb
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1 Answers1

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Yes, in fact the determinant $\det:(\mathbb R^n)^n\to \mathbb R$ is (up to constant multiple) the only alternating $n$-multilinear map (i.e. alternating $n$-covariant tensor, see John Lee, Introduction to Riemannian Manifolds, 2nd edition (2018), pages 400 and following). See for instance this question for a proof of this fact.

Maximilian Janisch
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  • Thanks! Would you be able to represent the determinant in a basis of simpler multilinear maps / tensors? (The second part of my question) And would that representation be different if you define the map on the row vectors of the matrix than the column vectors? – mihirb Jul 22 '20 at 22:43
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    If you choose the standard basis ${e^1,...,e^n}$ dual to ${e_1,...,e_n}$ then you can write it as the antisymmetrization of $e^1 \otimes e^2 ... \otimes e^n$ (potentially with some normalization factor), so yes to your first question. As for the second, if you let ${e_1,...,e_n}$ be the standard "row" basis and let ${e^1,...,e^n}$ be the dual basis, then you get the same "representation". – Osama Ghani Jul 23 '20 at 00:42
  • I’m sorry, but why is the determinant a (n-)covariant tensor instead of contravariant? Since it is the volume of the simplex, when vector components scale up, so does the det, right? Thanks – Bcpicao Sep 05 '23 at 19:18
  • @Bcpicao It is covariant because it's arguments are matrices, i.e. $n$ vectors in $\mathbb R^n$. If it were contravariant then its arguments would be $n$ covectors, i.e. $n$ elements of $(\mathbb R^n)^*$. And two your second question: Yes, the determinant is multilinear, that is, when you multiply any of its arguments by a constant, the result is also multiplied by the same constant. – Maximilian Janisch Sep 13 '23 at 07:16
  • Thank you for your quick response. I think I’m having a hard time seeing how the different equivalent definitions of tensors fit together in this case. In terms of multilinear maps, what you say about the arguments makes sense. But reasoning in terms of basis change, when we scale the basis vectors by 1/3, vector components get multiplied by 3, and the volume by 3^n, suggesting the determinant is contravariant. Can you help me resolve this apparent incongruency? Thank you. – Bcpicao Sep 13 '23 at 16:19
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    @Bcpicao I am not sure I understand your last sentences: When you multiply the arguments by a constant, say $3$, of an $n$-multilinear map, then the image is multiplied by, say, $3^n$. This is true for multilinear maps i.e. covariant tensors. Or am I missing something? – Maximilian Janisch Sep 15 '23 at 10:55
  • Im sorry, i caught my mistake, thanks! – Bcpicao Sep 15 '23 at 17:41