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 ...

12
votes

Accepted

### 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-...

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 ...

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 ...

11
votes

Accepted

### 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 $2$ ...

9
votes

Accepted

### 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 ...

9
votes

Accepted

### 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 (...

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 ...

8
votes

Accepted

### 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 ...

8
votes

Accepted

### 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 ...

8
votes

Accepted

### 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) ...

8
votes

Accepted

### 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 ...

7
votes

Accepted

### 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 ...

6
votes

Accepted

### 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 ...

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 ...

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 ...

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} \...

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 ...

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 ...

6
votes

Accepted

### 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_{...

5
votes

Accepted

### 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 ...

5
votes

Accepted

### 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 ...

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 ...

5
votes

Accepted

### 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,\...

5
votes

Accepted

### 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 ...

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.

5
votes

Accepted

### 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,\...

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 ...

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

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 ...

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