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2 votes
1 answer
35 views

Invariant theory for unitary groups $\mathcal{U}(n)$

I'm trying to understand the invariant theory of the unitary groups $\mathcal{U}(n)$ on tensor powers of their standard representations $V^{\otimes p} \otimes (V^*)^{\otimes q}$. Let $\mathcal{U}(n)$ ...
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-...
0 votes
1 answer
157 views

Can the derivative of eigenvectors with respect to its components be taken as zero if all eigenvalues are equal?

I want to ask a couple of follow up questions to the question answered on the thread "Derivative of eigenvectors of a matrix with respect to its components". I noticed that in the accepted ...
1 vote
0 answers
124 views

Can numerical differentiation be applied to tensor derivatives?

I know that for a 1D function, I can calculate the numerical derivative at every point, $\DeclareMathOperator{\d}{d\!} (x_1,y_1)$, with $\d y/\d x$ where $\d y = y_2 - y_0$ and $\d x = x_2 - x_0$. If ...
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 ...
1 vote
0 answers
124 views

Space of all orthogonal partially complex $2\times2\times2$ tensors

I am trying to understand the space of all orthogonal tensors, I asked a more general version of this question here but with no solution yet found. I want to look at the simplest case first, namely a $...
2 votes
1 answer
301 views

Best approximation with tensors of rank $\ge2$

Let $k\in\mathbb N$, $H_i$ be a (finite-dimensional, if necessary) $\mathbb R$-Hilbert space for $i\in I:=\{1,\ldots,k\}$, $H:=\bigotimes_{i\in I}H_i$ denote the tensor product$^1$ of $(H_i)_{i\in I}$ ...
4 votes
0 answers
126 views

Space of all orthogonal $2\times2\times2$ tensors

I am trying to understand the space of all orthogonal tensors, a question asked here before but with no real solution yet found. The solutions for order-$2$ tensors are clear so thus the simplest case ...
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 ...
2 votes
0 answers
63 views

Adjoint to "strict twocategory of strict twofunctors"

Let C be the category of strict twofunctors, featuring the addition of a Grothendieck universe. Strict twocategories are categories enriched over the category of categories. C has an internal hom ...
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\...
2 votes
1 answer
201 views

Combination of simple tensors - II

This is a follow-up question to Combination of simple tensors. I am interested in devising an alternative norm (I mean, other than the usual $\pi$ or $\epsilon$ norms) in the tensor product of two ...
2 votes
1 answer
228 views

Combination of simple tensors

I aksed this question on Math Stack Exchange 6 days ago, with no answer: https://math.stackexchange.com/q/4875445/1297919 Let $X$ and $Y$ two Banach spaces and let $X\otimes Y$ their tensor product. ...
5 votes
1 answer
463 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^{\...
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 \...
3 votes
1 answer
68 views

What is the best known bound for the bilinear complexity of $4\times 4$ matrices product

Assume we work on the complex field $\mathbb{C}$. And we use $\langle p,q,r\rangle$ to denote the bilinear complexity of product of a $p\times q$ matrix and a $q\times r$. Recently I read a paper on ...
0 votes
1 answer
104 views

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

Assume we work on any infinite field and 3-ordered tensor. Clearly for any tensor $T$, we have $\operatorname{srk}(T)\le \operatorname{rk}(T)$. Here, $\operatorname{srk}(T)$ (resp. $\operatorname{rk}(...
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 ...
1 vote
1 answer
142 views

Diagonalizability, orthogonal diagonalizability of higher order tensors and their being or not being dense in some suitable topology

For our discussion, we'll assume that we're working with $\mathbb{R}^m$ only, but much or all of the following discussion should be carried over immediately to any finite dimensional inner product ...
4 votes
0 answers
131 views

Is there a fast way to do this tensor power/trace operation?

Well, I asked this question on Math SE, and didn't get any responses, so I'm trying it here. Given an $M*N*P$ tensor $T$, is there a fast way of computing the following "eighth-power trace"? ...
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 ...
1 vote
0 answers
315 views

About "residual" scalar curvature in Einstein warped product manifold

I have an Einstein warped product manifold $M=B \times_f F$ with warped metric $g=g^B+f^2g^F$, where $F$ is Ricci flat, then its scalar curvature is $S^F=0$. It is well known that the scalar curvature ...
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) ...
1 vote
0 answers
155 views

Some kind of product of two 2d tensors to create a 3d tensor?

I recently need to apply the following concept of product of two 2d tensors to create a 3d tensor (tensors understood as generalized arrays): given two 2d tensors $A_{m\times n}$ and $B_{n\times p}$, ...
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 ...
2 votes
0 answers
43 views

$1$-parameter family of metrics preserving the normal direction

Let $(M^n,g)$ be a compact Riemannian manifold with boundary, $n \geq 2$, and let $N$ be the unit outward normal to $\partial M$. Denote by $S^2(M)$ the symmetric covariant $2$-tensors, by $S^2_0(M)$ ...
2 votes
0 answers
91 views

Is 'weak' Strassen Conjecture true?

$\newcommand{\rank}{\mathop{\mathrm{rank}}}$Strassen conjectured for two tensors $T_{1}$ and $T_{2}$, $\rank(T_{1}\oplus T_{2})=\rank(T_{1})+\rank(T_{2})$. This is not generally true according to ...
2 votes
3 answers
543 views

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

When the tensor algebra is presented, it is usually constructed as the direct sum of all tensor powers of the space. By this construction, the graded structure of the tensor algebra is easy to prove. ...
1 vote
0 answers
78 views

tensor dimension/reshaping group

Consider an $N$ dimensional tensor $T$ using the strided view representation used by PyTorch, i.e. we have a storage vector $S$ projected into $N$ dimensions using a size tuple $s$ and a stride tuple $...
1 vote
1 answer
205 views

Decomposition of tensor field on hypersurface

Let $(\mathcal{M},g)$ be a Lorentzian manifold, which is globally of the form $\mathcal{M}\cong I\times\Sigma$, where $I\subset\mathbb{R}$ ("time") and $\Sigma$ ("space") is some $...
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} =...
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} = - ...
0 votes
1 answer
70 views

Tensor nuclear norm for a binary 3rd-order tensor

I am interested in the low-rank approximation of a binary(01) third-order tensor. Does anyone know how to define its tensor nuclear norm based on whatever tensor decomposition methods? Could anyone ...
6 votes
1 answer
510 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_{...
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 ...
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 ...
2 votes
0 answers
319 views

What are examples of "perfect tensors"?

A "perfect tensor" is defined on the nLab very abstractly as "its tensor/hom-adjuncts $V^{\otimes k} \to V^{\otimes n - k}$ for $k \le n/2$ are isometries". The only example I'm ...
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 ...
1 vote
1 answer
103 views

What resource do Markov and Shi mean when they estimate tensor contraction complexity?

Markov and Shi in their paper Simulating quantum computation by contracting tensor networks define the contraction complexity as follows (page 10): The complexity of π is the maximum degree of a ...
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: $t_{ijkl} \rightarrow M_{ij, kl}$. ...
2 votes
1 answer
298 views

Decomposition of tensors

It is well known that every traceless, symmetric $2$-tensor can be decomposed uniquely into a Lie derivative part and a Codazzi part. Is there an analog for totally symmetric $k$-tensors?
1 vote
0 answers
66 views

Tucker decompositions over arbitrary fields

Given an $n$-mode tensor $\mathcal{T}\in\mathbb{R}^{d_1\times\dotsb\times d_n}$, there exists a Tucker decomposition of $\mathcal T$ of the form $$\mathcal{T} = \mathcal{X}\times_1 W_1\times_2\dotsb\...
2 votes
1 answer
1k views

Is there a geometric intepretation of the trace of tensor on a Riemannian manifold?

For a long time I thought the trace for matrices were just an elementary function with nice properties. But it is much more than that and really should be think of as a geometrical object (see ...
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 ...
-1 votes
1 answer
95 views

Multi-variable expansion of $e^{u^T X}$ as a power series in terms of tensors [closed]

I have the following question related to multivariable moment generating functions. Given two vectors $u,X\in\mathbb{R}^d$, I want to characterize the quantity $e^{u^T X}$ in terms of "powers&...
2 votes
0 answers
125 views

The Ricci curvature is bounded below by scalar curvature

So I have more questions coming from Dr Hamilton's "Three-Manifolds with Positive Ricci Curvature" paper here. I'm working in section 9 on Preserving Positive Ricci Curvature. In theorem 9.4,...
0 votes
0 answers
133 views

A question from Richard Hamilton's paper "A matrix Harnack estimate for the heat equation"

Richard Hamilton "A matrix Harnack estimate for the heat equation. Communications in Analysis and Geometry. 1(1993), 113-126." On page 125, at the end of the proof of Theorem 4.3, I abstract ...
0 votes
0 answers
199 views

Is the kernel $\vert d_X - d_Y \vert^p$ conditionally negative definite?

Given two finite metric spaces $(X,d_X)$ and $(Y,d_Y)$, for $p > 0$, define the kernel ($4$-D tensor) $K$ on $(X \times Y)^2$ by: $$K\big( (x_i, y_k), (x_j, y_l) \big) = \vert d_X(x_i, x_j) - d_Y(...
0 votes
0 answers
55 views

Solving nonlinear differential multi-variable equation with block-matrices

Here is the problem: Given a formula $f:\mathbb{R}^{n+k}\rightarrow\mathbb{R}^n$, written as $f(x_1(t),x_2(t),...,x_n(t);a1,...ak)$ with $k$ real unknown parameters $a_1,...,a_k$. For any $(a1,...ak)$,...
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 $$ \mathcal{A}=\mathcal{S} \times_1 U_1 ...\times_n U_n $$ ...