# Tag Info

### Does every ‘curvature’ tensor induce a metric?

Here's a quick summary. The answers provided in the link cited by @RBega2 have more details. Given a curvature-like tensor $R$ at a point, there always exists a metric whose curvature tensor at that ...
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### Non-tensor-representable ultrafilters on $\omega$

Recall that $\mathcal Z$ is a weak $P$-point if it is not in the closure of any countable subset of $\omega^* \setminus \{\mathcal Z\}$. A weak $P$-point is never the tensor product of two non-...
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### What is the largest tensor rank of $n \times n \times n$ tensor?

For tensors in $\mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3$ or in $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^3$, the maximum rank is $5$. See Bremner, Hu, On Kruskal's theorem ...
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### Why are matrices ubiquitous but hypermatrices rare?

As others have pointed out, higher-order tensors (i.e., hypermatrices) are in fact ubiquitous in mathematics, but they often aren’t discussed in detail as such because there’s not a lot you can say ...
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### Local diagonalisation of a degenerated 2d metric tensor

The answer is 'yes'. Here is how one can see this: Suppose that $g$ is a $(0,2)$ form on a neighborhood of the origin in the $xy$-plane such that the rank of $g$ is $1$ at the origin and $2$ ...
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### Is a flattening rank a lower bound for the border rank?

Yes, the flattening rank is a lower bound for border rank. First note that flattening rank is a lower bound for rank. If $T$ is a decomposable tensor (simple tensor, rank one tensor) then every ...
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### A different notion of a decomposable symmetric tensor (besides Veronese)

Your $\vee$ is essentially multiplication of polynomials. The variety of tensors $x_1 \vee \dotsb \vee x_m$ corresponds to polynomials that factor as products of linear factors. Points of the (...
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### What is expected (border) rank of the knonecker product of 3-tensors

We don't know. There are formats in which the equality is false for generic tensors: take $F^{n \times n \times 1}$ and $F^{n \times n \times n^2}$. Generic tensors in both formats are isomorphic to ...
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### higher order analogues of sylvester's law of inertia?

A generalization of Sylvester's classification of canonical quadratic forms (which is the "law of inertia") to cubic forms has been presented in Canonical forms for symmetric tensors (1984). The ...
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### Simple way to calculate the eigenvalues of a $2 \times 2 \times 2$ tensor

As explained in a previous MO question, there is no unique generalization of the eigenvalue of an $n\times n$ matrix to an $n\times n\times n$ tensor. One approach is to construct a higher-order ...
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### Bounds for metric in normal coordinate

As mentioned by Deane Yang in the comments and his (deleted) answer, one can estimate the components of the metric in normal coordinates using a transport ODE (I know it from Dolgov-Khriplovich (1983) ...
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### Is the asymptotic rank of a tensor bounded by (naive) border rank?

It is true that over $\mathbb{C}$ (and over every algebraically closed field) we have $\underline{R}^{\mathrm{Zariski}}(T) = \underline{R}^{\mathrm{original}}(T)$. The standard reference is the ...
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### Waring rank of monomials, and how it depends on the ground field

The answer to question 1 is affirmative. There are several lower bounds in various papers. I'll take the idea from https://arxiv.org/abs/1503.08253 (Buczyński and myself, "Some examples of forms ...
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### Vanishing of determinant of Cotton York tensor

The answer is 'no', the expression $\det(CY)$ does not vanish identically for metrics of the specified form. This follows by a direct computation, which is not all that difficult to do by hand, but ...
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### Hessian as a tensor, multi-dimensional taylor series, and generalizations

Sorry for reviving this question. Everything Tom said is correct, but there is more to say about "coordinate-free Taylor series". It is true that arbitrary jet bundles $J^k(M,N)$ are subtle. The ...
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### Quaternions as eigenvalues of rank 3 tensors

Because quaternions do not commute, there are two types of eigenvalues of an $n\times n$ quaternion matrix $A$: left eigenvalues solve $Av=\lambda v$ for some nonzero quaternion vector $v$, while ...
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### How to count the number of tensors over a finite field of tensor rank $r$?

This seems difficult. Deciding if a tensor has rank $\leq r$ (over a fixed finite field) is $\mathsf{NP}$-complete (Hastad, 1990). I haven't checked recently, but surely it's also the case that ...
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To your first question: no, even if we take out the orthonormal condition, by a dimension count argument (with one nontrivial exception). Assume there is generally. The $\lambda_i$ can be removed by ...