Questions tagged [tensor]

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4 answers
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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
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103 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
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
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20 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
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
16 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
10 votes
2 answers
9k 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
5 votes
1 answer
215 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
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3 votes
1 answer
388 views

Exactness of injective tensor products

For (algebraic) tensor products, it is well-known that the functor $A\otimes_R \cdot:Mod_R\rightarrow Mod_R$ is only (left-) exact when $A$ is a flat $R$-module. In particular, all vector spaces are ...
ABIM's user avatar
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2 votes
1 answer
210 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. ...
Lorenzo Guglielmi's user avatar
2 votes
1 answer
2k views

Eigenvalue and Eigenmatrix of a 3D Tensor - How to calculate it?

How to calculate easily the eigenmatrix of a 3D tensor. I try immersing the tensor in a big matrix, in my case, the tensor is of nxnxn and I can build an n^2 x n^2 matrix that contains all the "...
Alejandro Alvarez-Socorro's user avatar
1 vote
1 answer
2k views

Commutation of tensor products with inverse limits in a specific case

For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings)...
Duchamp Gérard H. E.'s user avatar