4
$\begingroup$

My background consists mostly of a good level in linear algebra, abstract algebra, undergrad calculus, topology & probability, and some working knowledge of geometric algebra and category theory. I'm currently learning differential geometry and tensor calculus, and hope to move onto geometric calculus and information geometry. However, I have a lot of trouble integrating abstract formulas and concepts if I don't have some intuitive mental model on which to rely.

My question is very broad: what is a coherent system of visualization for $(m,n)$-tensors ? Since this question can be answered in many ways, I'll be asking more specific questions as well. To help guide your answer (and hopefully help other learners like myself), here's some insight I've built up over the years on how to visualize some basic objects.

Scalars are simply single numbers from a field $F$. They are $(0,0)$-tensors. A scalar field can be viewed as a coloring over a manifold, where each point gets darker and redder as the scalar at that point tends towards positive infinity, whiter as it tends towards zero, and darker and bluer as it tends towards negative infinity.

Vectors are $(1,0)$-tensors. I understand and visualize them either as an oriented arrow, or as a "line with cyclic colors" (1D subspace with a repeated hue shift). The norm of your vector is equivalently the distance from the tail to the head of the arrow, or the distance to get from a specific shade to the next occurence of the exact same shade on your line. The order of the hue shift RGB or BGR encodes the orientation. This second picture is often useful in seeing geometric algebra as the algebra of subspaces of a vector space. Vectors, in 'matrix form', are column vectors ($n*1$ matrices). A vector field is a flow field where a particle follows the arrows, with one vector defined at each point on the manifold (like arrows for winds on a globe of the earth in the context meteorological data). A smooth vector field can also (sometimes? always?) be seen as a foliation, a space of solutions to a differential equation which divides the spaces into non-intersecting flow curves.

Covectors, 1-forms, $(0,1)$-tensors, I think I understand as well. By definition they are linear functions from $V$ to the underlying field $F$. They can be seen stacks of regularly spaced parallel (hyper)planes. They're your row vectors ($1*n$ matrices). They "eat" vectors by returning a scalar, operating somewhat like a dot product (for a euclidean metric). You can also see covectors as stacked (hyper)planes that hue-shift, and if the plane that passes through the origin is "pure red", the result of the dot-product-like operation (applying the linear form to a vector $v$) is the number of times our vector $v$ passes through "pure red" planes, plus whatever fraction of the way to the next red plane the arrowhead happens to lie on. A higher norm for a covector means that the sheets in the stack are closer together (ie, the same vector will pass through more sheets, returning a larger scalar). Covector fields are like topographical maps. Say a particle is going through the covector field: the more it goes "towards the top of the mountain", the denser the material it has to go through becomes, the more its speed is "eaten up" by the covector field; and vice-versa, it speeds up when "going down the mountain", because it's going from a dense medium to a fluid medium.

2-vectors, bivectors, $(2,0)$-tensors, are oriented areas. You can see them as parallelograms (formed as the outer product of two non-colinear vectors) or ellipses, or any 2D shape, so long as orientation (clockwise vs counter-clockwise rotation in the plane; or equivalently transparent vs textured side of the plane) and area (the norm of the bivector) are the same. I wonder if for this reason there's also a hue-shift-type visualization for bivectors; like representing a bivector as a specific "hue-shift tiling" of a full 2D subspace ? I can't quite work it out. Also, given that bivectors are $(2,0)$-tensors, am I correct to assume that you could make a "column vector of column vectors" that could "eat two successive 1-forms, or a single 2-form", as a "more appropriate" matrix form ? (By this latter question, I mean that I have trouble understanding how $(2,0)$-tensors, $(1,1)$-tensors and $(0,2)$-tensors should all look like rectangular matrices; it seems incoherent to me, like an abuse of notation: perhaps useful, but in need of clarification.)

As for 2-forms (which I guess one could also call cobivectors or bicovectors), I'd expect there be some intuitive version of "a shape, or repetition of parallel shapes, that eats bivectors" but I can't wrap my head around it. Part of my gut intuition from working on geometric algebra, which associates covectors to $(n-1)$-vectors, and associates applying a covector to a vector to obtaining a pseudoscalar value from the product of an $(n-1)$-vector and a $1$-vector which are independent (though I have no idea if this analogy is legitimate) tells me I'd have to take a stack of $(n-2)$-dimensional spaces to represent a 2-form. Would this mean that if I consider a 1D linear subspace of $R^3$ and the set of all affine spaces parallel to this line (like a bundle of straws stretching out infinitely), a "closeness" or density of each straw to its neighbors would somehow encode the intensity of a flow through a bivector, and thus a 2-form ? Given that they are (0,2)-tensors, am I correct to assume that they could be represented as "a line vector of line vectors" ? And if this image of 2-forms is correct, how does this relate to the idea of metric tensors ?

Linear maps from $F^n \to F^m$ are $(m*n)$-matrices. They can be understood as mapping $(1,0)$-tensors of dimension $n$ to $(1,0)$-tensors of dimension $m$ (by right multiplication with a vector), so I suppose they are $(1,1)$-tensors (as in, the covariant part of the matrix is combined with the contravariant column vector, and only the contravariant part of the matrix/(1,1)-tensor remains). This is also coherent with left multiplication of an $(m*n)$-matrix by an $m$-dimensional covector (like in the context of Markov chains), as in it consumes an $m$-dimensional $(0,1)$-tensor and returns an $n$-dimensional $(0,1)$-tensor.

Finally, you often see generalized dot products (bilinear symetric forms) being used with the pattern "$x^TSy$", with $S$ a symmetric matrix. Am I correct to assume that they are $(1,1)$-tensors which are simultaneously fed $(1,0)$-tensor and a $(0,1)$-tensor ? Or are they indeed like "metric tensors", and thus (0,2)-tensors ?

Given what you've just read, are there any glaring mistakes in this picture ? Are there useful visualization and insights you care to share that might go well with the current picture (both to illustrate intuitive and counter-intuitive ideas) ? Visualization tools or models for higher-valence mixed tensors ? For tensor fields (other than the basic "bundle = hairy manifold where each hair is a tensor") ? Of the idea of vectors, covectors, etc, $(m,n)$-tensors in infinite-dimensional vector spaces (which I haven't even approached here) ? Any insight on seeing tensor operations as operations on hypermatrices ? All insight from any branch that touches upon these themes is welcome !

I'm sorry if this makes a lot of diffuse questions, it's just that I'm having trouble making the whole picture of "$(m,n)$-tensors, their geometry, their algebraic representation and how one computes with them" all coherent in my mind.

Thanks for reading, and for your help !

$\endgroup$
14
  • 1
    $\begingroup$ I would like you take a hike or "promenade" along my answers on the subject you may find Ad hoc. For example on math.stackexchange.com/search?q=user:74166+[tensors], in math.stackexchange.com/… or in math.stackexchange.com/search?q=user:74166+[differential-forms]. $\endgroup$
    – janmarqz
    Nov 24, 2019 at 16:06
  • 1
    $\begingroup$ Thanks, I'll definitely check them out and keep you posted if I have questions (possibly on the answers themselves) ! :) $\endgroup$ Nov 24, 2019 at 16:15
  • 1
    $\begingroup$ I will be thinking in an answer to your question $\endgroup$
    – janmarqz
    Nov 24, 2019 at 16:39
  • 1
    $\begingroup$ The tensor of certain rank form vector spaces. The tensor spaces of rank two, $(2,0)$, $(1,1)$ and $(0,2)$ are isomorphic among them, and the operation of contracting them against the metric tensor $g_{ij}$ or $g^{ij}$ serve to define these isomorphisms. All elements of these 3 vector spaces can be described by bi-indexed quantities, so they are square matrices that include symmetric ones but an arbitrary one no necessarily must be nor can be transformed to a symmetric one. $\endgroup$
    – janmarqz
    Nov 24, 2019 at 22:45
  • 1
    $\begingroup$ That's quite alright, thank you for your time, I'll keep reading your posts and others in the meantime :) $\endgroup$ Nov 25, 2019 at 13:25

1 Answer 1

1
$\begingroup$

Semi-answer based on the comments below.

One ought to think tensor as:

  1. They are vectors in different spaces. For the tensor product, we exemplify by taking $w=w_s\beta^s$ a one form $(1,0)$ and one says that $w_s$ are its components which could be functions on the coordinates, and if $A=A_{ij}$ is a two covariant tensor, $(2,0)$, in such a way that $A=A_{ij}\beta^i\otimes\beta^j$ is used to indicate its precedence, then their tensor product $$w\otimes A$$ would be a vector on the space $(3,0)$ of tensors and should have the components $$(w\otimes A)_{ijk}=w_iA_{ij}$$ that is, the new tensor $w\otimes A$ has components by multiplying those scalar functions: the components of $w$ with the components of $A$.

  2. Contracting. One also could construct the associated tensor $\bar w$ whose components are $\bar w^k=g^{ks}w_s$, where $g^{rt}=\beta^r\bullet\beta^t$ are the entries of the metric co-tensor. With $\bar w$ two types of contractions are possible, and give two $(1,0)$-tensors, each with components: $$(wA)_i=\bar w^sA_{is};$$ or $$(wA)_j=\bar w^s A_{sj},$$ in both using the $s$-sum Einstein's Convention.

  3. Symmetry. If for $A$ you take one with a symmetric matrix, those two types are equal.

  4. For a $(1,0)$ and a $(1,1)$ you would have $w_s$ and $B^t{{}}_u$ respectively, and you could contract them as $$w_sB^s{{}}_u,$$ to get the $u$-th component of a $(1,0)$ tensor or contract as $$\bar w^sB^t{{}}_s,$$ for the $t$-th component of another $(0,1)$ tensor.

$\endgroup$
5
  • $\begingroup$ Thanks, your examples clear out some of the problems I had with the algebraic part (Einstein notation summations) of tensors. A question: why the bullet operator in 2) ? $\endgroup$ Nov 26, 2019 at 10:58
  • $\begingroup$ that is the way one get the metric from the base covectors, that also code the length and angles between them. The $g^{ij}$ are use to construct all kind of associated tensors of any rank $\endgroup$
    – janmarqz
    Nov 26, 2019 at 13:55
  • $\begingroup$ What I don't get is: why isn't it a normal tensor product or outer product ? I've never seen this bullet product before $\endgroup$ Nov 26, 2019 at 15:19
  • $\begingroup$ to symbolize an inner product among the betas, and damright! is a symbol rarely employt $\endgroup$
    – janmarqz
    Nov 26, 2019 at 16:10
  • 1
    $\begingroup$ I see, thanks for clearing that up ! $\endgroup$ Nov 26, 2019 at 16:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .