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109 votes
15 answers
12k views

Why are matrices ubiquitous but hypermatrices rare?

I am puzzled by the amazing utility and therefore ubiquity of two-dimensional matrices in comparison to the relative paucity of multidimensional arrays of numbers, hypermatrices. Of course ...
Joseph O'Rourke's user avatar
28 votes
4 answers
6k views

Is there any way to rewrite a partial differential equation using language of differential forms, tensors, etc?

My question is: usually, a partial differential equation, for example, those coming from physics, is written in a language of vector calculus in a local coordinate. Is there any way (or any algorithm) ...
HYYY's user avatar
  • 1,499
21 votes
3 answers
5k views

Geometrical meaning of the Ricci Tensor and its Symmetry

Let $M$ be a smooth, pseudo-Riemannian manifold with $\dim(M) \ge 2.$ Let $\nabla$ be any affine connection on $M$. No reason for it to be the Levi-Civita connection. All we assume is that it has zero ...
Fly by Night's user avatar
20 votes
3 answers
9k views

Hessian as a tensor, multi-dimensional taylor series, and generalizations

The Hessian matrix $\{\partial_i \partial_j f \}$ of a function $f:\mathbb{R}^n \to \mathbb{R}$ depends on the coordinate system you choose. If $x_1,\cdots,x_n$ and $y_1,\cdots,y_n$ are two sets of ...
David Corwin's user avatar
  • 15.4k
19 votes
3 answers
2k views

Best known bounds on tensor rank of matrix multiplication of 3×3 matrices

Years ago I attended a conference where they taught us that matrix multiplication can be represented by a tensor. The rank of the tensor is important, because putting it into minimal rank form ...
Brian Rushton's user avatar
19 votes
4 answers
3k views

Equations satisfied by the Riemann curvature tensor

It is well known that the Riemann curvature tensor of a metric satisfies \begin{eqnarray} R_{jikl}=-R_{ijkl}=R_{ijlk},(1)\\ R_{klij}=R_{ijkl},(2)\\ R_{i[jkl]}=0 \mbox{(1st Bianchi identity)}.(3) \end{...
asv's user avatar
  • 21.8k
19 votes
2 answers
1k views

Exponentiation of vector spaces?

This question occurred to me while thinking on another one here, Name for an operation on matrices? Can one define in an invariant way a binary operation on finite-dimensional vector spaces - let us ...
მამუკა ჯიბლაძე's user avatar
18 votes
2 answers
2k views

What is the largest tensor rank of $n \times n \times n$ tensor?

The tensor rank of a three dimensional array $M[i,j,k], i,j,k\in [1,\ldots,n]$ is the minimal number of vectors $x_i,y_i,z_i$, such that $M=\sum_{i=1}^d x_i\otimes y_i\otimes z_i$. From dimension ...
Klim Efremenko's user avatar
16 votes
2 answers
3k views

Who coined the name tensor and why?

Who coined the name "tensor" and why? What does the word "tensor" really mean, not the mathematical definition?
Meseret Tiruye's user avatar
11 votes
2 answers
10k views

Derivative of eigenvectors of a matrix with respect to its components

Suppose that $B$ is a real, positive-definitive symmetric ($3\times3$) matrix (more accurately, $B$ is a tensor) with distinct eigenvalues, and that we can write it as $$ B= \sum_{i=1}^3 \lambda_{i}(...
Jeff Tehrani's user avatar
10 votes
2 answers
1k views

Does the hyperdeterminant calculate a quantity akin to the volume of a parallelepiped?

If $M$ is an $n \times n$ matrix, $|\det(M)|$ is the volume of the $n$-dimensional parallelepiped spanned by the column vectors of $M$.                  ...
Joseph O'Rourke's user avatar
9 votes
1 answer
326 views

Geometric Interpretation of Multiplication in Pure Cubic Number Fields and Beyond

I got interested in the question of possible geometric interpretations of the multiplication in algebraic number fields of degree $>2$ (with application to multiplication of units in the ring of ...
Andreas Rüdinger's user avatar
9 votes
2 answers
2k views

A generalization of Boolean matrix multiplication for order-3 tensors

The Boolean matrix product of two 0-1 $n \times n$ matrices $A$ and $B$ is the matrix $C$ defined as $$C[i,j] = \vee_{k=1}^n (A[i,k] \wedge B[k,j]).$$ If $A = B$ and the matrix is an adjacency matrix ...
Ryan Williams's user avatar
8 votes
1 answer
729 views

Recovering a smooth manifold from its tensor fields

1) Consider the algebra $\mathcal T (M) = \bigoplus \limits _{p, q \ge 0} \mathcal T ^{p, q} (M)$, where $\mathcal T ^{p, q} (M)$ is the space of tensor fields of type $(p,q)$. I should endow it with ...
Alex M.'s user avatar
  • 5,407
8 votes
1 answer
238 views

Question on Nash's paper on $C^1$ isometric immersions: Why approximating the error tensor $\delta$?

I am trying to go through the classical paper by Nash on the existance of $C^1$ isometric immersion of a Riemannian manifold $(M,g)$ (here is the Jstor link: https://www.jstor.org/stable/1969840?seq=1#...
Nick A.'s user avatar
  • 203
8 votes
0 answers
267 views

Generalization of a standard algebraic group theory result for a tensor problem

$\DeclareMathOperator\GL{GL}$Let $X$, $Y$, $Z$ be $\mathbb{C}$-vector spaces, and let $A\subseteq X$ and $B\subseteq Y$ and $C\subseteq Z$ be linear subspaces. Let $V=X \otimes Y \otimes Z$, acted on ...
Ben's user avatar
  • 980
7 votes
1 answer
192 views

Non-tensor-representable ultrafilters on $\omega$

If ${\cal U}$ and ${\cal V}$ are ultrafilters on non-empty sets $A$ and $B$ respectively, then the tensor product ${\cal U}\otimes{\cal V}$ is the following ultrafilter on $A\times B$: $$\big\{X\...
Dominic van der Zypen's user avatar
7 votes
1 answer
231 views

Is $\max_{\|x\|_p=\|y\|_p=1} |\langle x, Ay\rangle|$ equivalent to $\max_{\|x\|_p=|} |\langle x, Ax\rangle|$ for symmetric $A$ & $p\geq 2$?

Let $A\in \mathbb{R}^{n\times n}$ be a symmetric matrix, and consider the $l_p$ norm ($p\geq 2$). Can we prove that the following problems are equivalent: $$\max_{\|x\|_p=\|y\|_p=1} \left| \langle x, ...
jayki's user avatar
  • 135
7 votes
2 answers
440 views

How to count the number of tensors over a finite field of tensor rank $r$?

For simplicity, work over $\mathbb F_2$ and only consider order-$3$ equilateral tensors. For $r\in\mathbb Z_{>0}$, how many tensors $\mathfrak{T}\in\mathsf{Ten}_{n}^{\otimes 3}(\mathbb F_2)$ are ...
Yihan Zhang's user avatar
7 votes
1 answer
356 views

Is there any sort of higher-order SVD (quadratic and above) for dimensionality reduction?

(Posted this on math.stackexchange and cross.correlated over more than a week ago, but didn't get an answer, and this is a question in my research so this seems like it might have been the better ...
user650261's user avatar
6 votes
1 answer
766 views

Does every ‘curvature’ tensor induce a metric? [duplicate]

So we know that given a Riemannian manifold $(M,g)$ we can calculate its Riemannian curvature $R$. This covariant $4$-tensor field then satisfies some important symmetries \begin{gather*} R_{ijkl} = - ...
Matthew Lou's user avatar
6 votes
1 answer
511 views

Waring rank of monomials, and how it depends on the ground field

The Waring rank of a degree-$d$ homogeneous polynomial $p$ is the least integer $r$ such that you can write $p$ as a linear combination of $r$ $d$-th powers of linear forms $\{\ell_k\}$: $$ p = \sum_{...
Nathaniel Johnston's user avatar
6 votes
1 answer
2k views

Is there a relation between the Hessian matrix and the structure Tensor?

for a 2 dimensional image, i am interested to find a relation between the Hessian $H = \begin{pmatrix} \frac{\partial^2 I}{\partial x^2}&\frac{\partial^2 I}{\partial x \partial y}\\ \frac{\partial^...
R3tsn0Om's user avatar
6 votes
1 answer
621 views

Tensor matricizations and their decompositions

Suppose we have a 4-index tensor $t_{ijkl}$ (all 4 dimensions are equal size). We can make a matrix out of it by taking first and last two indexes as new indexes: $t_{ijkl} \rightarrow M_{ij, kl}$. ...
qbit-'s user avatar
  • 71
5 votes
2 answers
425 views

Local diagonalisation of a degenerated 2d metric tensor

Consider a smooth 2d-manifold $M$ and let $g$ be a smooth $(0,2)$-tensor satisfying $rk(g)\geq1$ everywhere. Obviously if $rk(g)=2$ at a point $p\in M$ then $g$ is locally diagonalisable (i.e. there ...
Athena's user avatar
  • 275
5 votes
2 answers
339 views

Example of a curvature with no associated metric

Is there a concrete example of a $4$ tensor $R_{ijkl}$ with the same symmetries as the Riemannian curvature tensor, i.e. \begin{gather*} R_{ijkl} = - R_{ijlk},\quad R_{ijkl} = R_{jikl},\quad R_{ijkl} =...
Matthew Lou's user avatar
5 votes
2 answers
572 views

Bounds for metric in normal coordinate

Let $M$ be a Riemannian $n$-manifold and $x \in M$. For the metric tensor $g_{ij}$ of geodesic normal coordinates at $x$, there is a formula $g_{ij}(y) = \delta_{ij} + \frac13 R_{kijl} y^k y^l + O(\|y\...
Uzu Lim's user avatar
  • 903
5 votes
1 answer
186 views

What is expected (border) rank of the knonecker product of 3-tensors

Given two three order tensors $T$ and $S$ in $F^{m\times n\times p}$ and $F^{a\times b\times c}$. Clearly $\operatorname{rk}(T\otimes S)\le \operatorname{rk}(T)\operatorname{rk}(S)$. Does the equality ...
Nick Chen's user avatar
  • 151
5 votes
1 answer
610 views

Is a flattening rank a lower bound for the border rank?

Suppose $T \in V_1 \otimes \cdots \otimes V_k$ is a tensor, where each $V_i$ is a finite dimensional complex vector space. A $1$-flattening (or a flattening) is a realization of $T$ as a matrix in the ...
SMD's user avatar
  • 500
5 votes
1 answer
241 views

Is the asymptotic rank of a tensor bounded by (naive) border rank?

Write $R(T)$ for the rank of an order-$3$ tensor $T \in \mathbb C^{m \times n \times p}$ over the complex numbers. If $T' \in \mathbb {C}^{m' \times n' \times p'}$ is another such tensor then let $T \...
Sean Eberhard's user avatar
5 votes
1 answer
473 views

higher order analogues of sylvester's law of inertia?

Sylvester's law of inertia (here I quote wikipedia) If A is the symmetric matrix that defines the quadratic form, and S is any invertible matrix such that D = SAS^{T} is diagonal, then the number ...
mathstudent42's user avatar
5 votes
1 answer
5k views

Levi-Civita symbol

Is the Levi-Civita symbol a tensor? R. A. Sharipov afirm (In "Quick Introduction to Tensor Analysis", page 30) that "...the Levi-Civita symbol is NOT a tensor..." $\epsilon_{jkq}=\epsilon^{jkq}=\...
Esteban's user avatar
  • 53
5 votes
1 answer
464 views

Bochner Laplacian in coordinates

Sorry if this is a too basic question, but I didn't find an answer anywhere: The connection Laplacian, or Bochner Laplacian, is the differential operator acting on $k$-tensor fields $T\in\Gamma^{\...
B.Hueber's user avatar
  • 1,171
5 votes
1 answer
219 views

Tell me something about these "component tensor" TQFT's

I noticed that there is a class of TQFT's that exists for every dimension $n\geq1$. It's probably well-known because it's quite simple, but I'm looking for a standard name or a better way to think ...
David Spivak's user avatar
  • 8,659
5 votes
1 answer
5k views

Is there a generalization of eigenvalues and eigenvectors to tensors?

Two perhaps ill-posed or just silly questions: Let $n>0$, $T$ be an $(n+2)$-tensor, and $\otimes$ denote the Kronecker product of tensors. Is there a tensor generalization for the fundamental ...
hypnotoad's user avatar
5 votes
1 answer
309 views

Looking for a tractable algorithm or formula for the determinant of a tensor

It is possible to define the determinant of a tensor. We think of a tensor as a collection of numbers but this collection easily extends to a proper multilinear map. If $T:\{1,....,n\}^m\to \mathbb C$ ...
Pete L.'s user avatar
  • 183
5 votes
0 answers
101 views

How is this product of tensors defined?

I am reading the paper “ The first eigenvalue of a small geodesic ball in a riemannian manifold”, by Karp and Pinsky, from where I took the following: Here, $\Delta_{-2}$ denotes the usual Laplacian ...
Eduardo Longa's user avatar
4 votes
1 answer
547 views

Question on Lorentzian geometry

I apologize in advance if this is a too basic question. Let $(M,g)$ be a Lorentzian manifold with signature convention $(-,+,\dots,+)$. Now, lets suppose $X\in\Gamma(TM)$ defines a global time-...
B.Hueber's user avatar
  • 1,171
4 votes
1 answer
432 views

Simple way to calculate the eigenvalues of a $2 \times 2 \times 2$ tensor

I am working with hypergraphs. The various matrices associated with hypergraphs are hypermatrix or tensors. I am interested in spectral aspects. In particular, I want to find all the eigenvalues ...
GA316's user avatar
  • 1,269
4 votes
1 answer
544 views

Are constant connection coefficients uniquely determined by the (1,3) curvature coefficients?

Suppose that on a certain coordinate system the coefficients $\Gamma^i_{jk}$, $i,j,k=1,\cdots, n$, of a linear connection are constant. We do not require compatibility with a metric, however I am ...
Ettore Minguzzi's user avatar
4 votes
1 answer
981 views

Symmetric tensor decomposition

Let $T$ be an order-$k$, rank-$m$ symmetric tensor, that is, $T=\sum_{j=1}^m v_j\otimes v_j \otimes \cdots \otimes v_j$, where the Segre outer product is taken $k$ times, with $v_j\in\mathbb{R}^d$ for ...
hookah's user avatar
  • 1,096
4 votes
1 answer
3k views

Taylor's series for Lie groups

Let $G_1$ and $G_2$ be two (matrix) Lie groups, with $L(G_1)$ and $L(G_2)$ their respective Lie algebras. I am interested to know if there is a well developed theory to approximate a (sufficiently) ...
Alessandro Saccon's user avatar
4 votes
1 answer
747 views

The Laplacian of an expression involving the Ricci tensor

While doing some computations on a compact Riemannian manifold I have reached the following expression: $$ \Delta_y \big( Ric_y (\exp_y ^{-1} x, \exp_y ^{-1} x) \big) (x)$$ where $\Delta_y$ is the ...
Alex M.'s user avatar
  • 5,407
4 votes
2 answers
745 views

Indexed tensor manipulation CAS

hello. I am looking for tensor manipulation software that would allow me: declare indices declare results of contraction (or simplification rules) allow algebraic simplifications and expansion index ...
user5925's user avatar
4 votes
1 answer
212 views

Singular value decomposition for tensor

I am looking at (the limitation of) the extension of the singular value decomposition to tensors. I would like to show that there is a tensor $A_{i,j,k}$ that cannot be decomposed in the following ...
Hans's user avatar
  • 2,239
4 votes
2 answers
433 views

What is the current status of representation theory of $n$-ary groups in terms of hypermatrices?

An $n$-ary group is a generalization of the usual concept of a group where the binary operation (2-argument operation) is instead an $n$-ary ($n$-argument) operation. More info here on Wikipedia. I ...
Sidharth Ghoshal's user avatar
4 votes
1 answer
2k views

Representation theory of (anti)self-dual tensors

I am using usual physics notations and I guess the physics motivations of this question are obvious. Let a basis of the $SO(n,m)$ Lie algebra be denoted by $S^{\mu \nu}$ and the Lie algebra be, $[S^{...
Anirbit's user avatar
  • 3,541
4 votes
1 answer
119 views

Construct a 4th-order tensor with matricization ranks $r$ that is not rank $r$

I ask you for this possibly not so simple task: Explicitly construct a 4th-order tensor $A \in \mathbb{C}^{n_1 \times \ldots \times n_4}$ that does not have (border) rank $r$, but for which each ...
Sebastian K.'s user avatar
4 votes
1 answer
494 views

Characterization of all-orthogonal tensors

In the paper [1], it is proven in Theorem 2 that any $n$-tensor $\mathcal{A}\in\mathbb{R}^{d_1\times...\times d_n}$ can be decomposed as $$ \mathcal{A}=\mathcal{S} \times_1 U_1 ...\times_n U_n $$ ...
Bonnevie's user avatar
4 votes
1 answer
1k views

Mathematica package for supergravity and string theory

I am looking for a Mathematica package that can manipulate tensors for supergravity, string theory or M-theory. I am particularly looking for a package that can do spinor and Clifford algebra ...
Kevin Ye's user avatar
  • 367