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15 votes

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 ...
Deane Yang's user avatar
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12 votes
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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-...
Will Brian's user avatar
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12 votes

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 ...
Zach Teitler's user avatar
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11 votes

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 ...
Henry Cohn's user avatar
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11 votes
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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 $2$ ...
Robert Bryant's user avatar
9 votes
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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 ...
Zach Teitler's user avatar
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9 votes
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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 (...
Zach Teitler's user avatar
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9 votes

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 ...
Vladimir Lysikov's user avatar
8 votes
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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 ...
Carlo Beenakker's user avatar
8 votes
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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 ...
Carlo Beenakker's user avatar
8 votes
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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) ...
Igor Khavkine's user avatar
8 votes
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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 ...
Vladimir Lysikov's user avatar
7 votes
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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 ...
Zach Teitler's user avatar
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6 votes
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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 ...
Robert Bryant's user avatar
6 votes

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 ...
Daniel Barter's user avatar
6 votes

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 ...
Carlo Beenakker's user avatar
6 votes

Can the eigenvalues of a real symmetric tensor be complex?

Let us take $n=2$. Let $T_{112} = T_{121} = T_{211} = 1$, $T_{222} = \frac{43}{9}$ and $T_{ijk} = 0$ otherwise. Consider the vector $\mathbf{x} = \left( \begin{array}{c} \frac{5}{4} \\ i \frac{3}{4} \...
Malkoun's user avatar
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6 votes

Proving the graded structure of the tensor algebra from only the universal property

$\newcommand\T{\mathrm T}$I assume that "the universal property" means that, for every $R$-algebra $S$, every $R$-module map $V \to S$ extends uniquely to an $R$-algebra map $\T V \to S$. In ...
LSpice's user avatar
  • 12k
6 votes

Example of a curvature with no associated metric

@AntonPetrunin's comment points to, I think, another way to describe the counterexample given by Robert Bryant in his answer. Consider a curvature-like tensor $$ R_{ijkl}(dy^i\wedge dy^j)(dy^k\wedge ...
Deane Yang's user avatar
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6 votes
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Example of a curvature with no associated metric

A simple example (which just uses Deane Yang/Robert Bryant's idea) is to consider any space of dimension at least three and consider the tensor field $$ R_{ijkl} = f(x)(\delta_{ik}\delta_{jl}-\delta_{...
Gabe K's user avatar
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5 votes
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Derivative of eigenvalues w.r.t. a tensor

Let $p(\lambda)$ be the characteristic polynomial $p(\lambda)=\det(E-\lambda I)$. Then $p(\lambda)=(\lambda_1(E)-\lambda)(\lambda_2(E)-\lambda)(\lambda_3(E)-\lambda)$. Differentiate in $E$ and then ...
Ben McKay's user avatar
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5 votes
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Tell me something about these "component tensor" TQFT's

The theory you describe is Dijkgraaf-Witten theory with target space a discrete set with $r$ elements. In general, if $X$ is a $\pi$-finite space (i.e., a space with finite homotopy groups, all but ...
Yonatan Harpaz's user avatar
5 votes

Why are matrices ubiquitous but hypermatrices rare?

This answer is relatively simpleton but I think it gets to the heart of the matter. Groups are very ubiquitious in mathematics. More so than non binary n-ary groups. Wherever there are/can be groups ...
Sidharth Ghoshal's user avatar
5 votes
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Eigenvalue and Eigenmatrix of a 3D Tensor - How to calculate it?

You ask for the eigenvalues of an $m=3$-order $n$-dimensional tensor $M$. There is no unique definition of the "eigenvalue" $\lambda$ for $m\geq 3$. One frequently used definition is $$\sum_{i_2,i_3,\...
Carlo Beenakker's user avatar
5 votes
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Symmetric tensor decomposition

A recent introduction is Carlini, et al, Four lectures on secant varieties. Adam mentioned Landsberg, Tensors: Geometry and Applications. In brief: 1(a). If $T$ is a symmetric tensor of tensor rank ...
Zach Teitler's user avatar
  • 6,227
5 votes

How far is the slice rank of a tensor from its CP rank

No. For $n \times n \times n$ tensors slice rank does not exceed $n$, while tensor rank can be as large as $\frac{n^3}{3n - 2} \sim \frac{n^2}{3}$ by dimension count.
Vladimir Lysikov's user avatar
5 votes
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Bochner Laplacian in coordinates

Example 10.1.32 (which starts on page 456) does not consider $\nabla$ the Levi-Civita for a Riemannian metric. It is considering a general vector bundle $E$ equipped with a Hermitian metric $\langle,\...
Willie Wong's user avatar
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4 votes

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 ...
Joshua Grochow's user avatar
4 votes

Is there a generalization of eigenvalues and eigenvectors to tensors?

There are several generalizations of the concept of "eigenvalues" to tensors. A good starting point on this topic is this paper by L.-H. Lim
Suvrit's user avatar
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4 votes

A kind of "Curvature tensor" for higher dimensional tensors

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 ...
user44191's user avatar
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