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Every time I get hear about tensors: "It's just a multidimensional array!" Ok, simple enough. But then I go to wikipedia get hit with differential geometry, m

• krastanov 11y ago
  Tensors are most definitely not just multidimensional arrays. (this is a pet peeve of mine) Tensors have geometric meaning just like vectors (just like vectors are not just single-axis arrays). It would be too long to delve into the geometry (especially if it is abstract "geometry" like the geometrical space spanned by the songs in the Pandora database or the space spanned by the faces in a face-recognition software). Instead I will try to give an example: You all know what rotating a vector means. All the components change in certain way, but the geometrical meaning of this object is preserved in some sense. This is not true for some arbitrary list of numbers. Well, a tensor is just a more complicated geometrical structure that still has very specific rules governing what its components can be and how they can change. And a counterexample: the Christoffel Symbol is a multi-dimensional array that is not a tensor. P.S. It is true that it is a very common abuse of nomenclature to call all multidimensional arrays "tensors".
  • chestervonwinch 11y ago
    It's true though that vectors are basically 1D lists if we're speaking about elements from the vector space R^n. Is it true that tensors are multidimensional arrays if they come from the space R^{m1 x m2 x ... x mn} ? I'm also confused: do you think about a tensor as a vector from vector space? as a object that maps between vector spaces? or something completely different? I realize there's a LOT of detail here - just trying to flesh out my 500,000,000 mile view a little more.
    • tel 11y ago
      It's not true, though. A list of numbers in R is not a vector until you imbue the position of each of those numbers with a geometric basis, or, alternatively, relate that list of numbers to the set of all lists of such numbers along with some algebraic operations which turn it into a vector space. This is similar with tensors. It's even harder to talk about them, though, without the geometric/algebraic bits.
    • im3w1l 11y ago
      They use tensors here to mean simply multidimensional array. And as this discussion shows, that choice of words is unfortunate and confusing.
      • rspeer 11y ago
        What I'm seeing is that physicists and geometers disagree with programmers about what "tensor" means -- not in that their views are inconsistent, but in that the programmers' view is too limited for the things that the physicists and geometers want to do with tensors. They also disagree about what "vector" and "matrix" mean for the same reason, but they can't really put that horse back in the barn. But "tensors" remain unfamiliar enough that this disagreement can come up whenever anyone says "tensor". Here's a possible truce: - Programmers should default to talking about "arrays" - Geometers and physicists should default to talking about "vectors", "matrices", and "tensors" - We all generally recognize that some (vectors|matrices|tensors) are arrays, and some arrays are (vectors|matrices|tensors), and it's not the worst thing ever if someone uses one to mean the other - We all generally recognize that you can sometimes do damn cool things by using an array as a tensor or a tensor as an array, so there isn't that much benefit in shouting "NO THEY'RE TOTALLY DIFFERENT".
        • tel 11y ago
          Eh, I think as you stack more and more "structure" atop something it becomes less and less appropriate to confuse it with its representation. There is afterall already a completely appropriate, cross-disciplinary term for what "programmers" here call "tensors": multidimensional arrays, or even nd-arrays. No mathematician will bat an eye if you say that a tensor can be represented by an nd-array---but if you equate them, then you're clearly missing something.
  • evanpw 11y ago
    I'm not sure if I'm disagreeing with you, but I really dislike the "collection of numbers that transforms in this way" definition of vectors, tensors, etc. I get why it's useful in calculations (that's why physicists like it), but it seems really inelegant to me. To me, a tensor is a function that maps a collection of m vectors to a collection of n vectors, such that every output is linear in each input. It's true that if you choose a basis, then you can write down an array of numbers which identifies the function, and that changing the basis causes those numbers to change in a certain way, but that's not what a tensor "is". Of course, the beauty of mathematics is that there are many different ways of looking at the same object, so physicists can deal with their arrays of numbers, I can play with my multilinear functions, and everyone gets the same answer when we ask the same question.
    • cygx 11y ago
      I really dislike the "collection of numbers that transforms in this way" definition of vectors, tensors, etc. It is not totally without merit if you consider the tensor bundle as associated to the principal bundle of linear frames. However, that's hardly suitable as an introduction to the topic.
      • tel 11y ago
        It's probably the Wikipedia introduction, though...
  • Enzolangellotti 11y ago
    So, a tensor is kind of a Rubik Cube shaped matrix?
    • ajkjk 11y ago
      A rank-3 tensor is. It's any number of dimensions you need.
• tel 11y ago
  If you're familiar with quadratic forms in multidimensional calculus you'll know that any bilinear relationship f(x, y) on two vectors from two vector spaces x and y can be represented by a matrix f(x, y) = x' A y for some matrix A. A tensor generalizes this idea to multilinear forms f(x, y, z, w, q) where if you play with the basic idea you'll probably realize that if such an operation were to be represented by some kind of "matrix" A then that matrix would have to be multidimensional. This kind of multidimensional array is a "tensor" although the theory gets a little more complex than just that. Why isn't any old multidimensional array a tensor? Because we want to see them as arising not as just an arbitrary collection of numbers but instead as representatives of these kinds of multilinear mappings. In particular, we want to know that tensors are invariant under changes of bases. This is similar to how we like to think of matrices as representing "some" linear mapping, but there might be many changes of bases in the source and target vector spaces which change the actual numbers inside without changing the ultimate mapping itself. Ultimately, to get the fullest understanding of this we would need to talk about vector dual spaces and note that expressions like f(x, y, z, w, q) could involve two "types" of vector spaces, covariant and contravariant, which have different kinds of transformation-under-basis-change properties. To truly be a tensor your multi-dimensional matrix must understand and respect all of these kinds of invariances.
  • defen 11y ago
    Is it fruitful to think of upper/lower indices as a type system for linear transformations?
    • tel 11y ago
      I definitely think so. I think it's done informally regardless, but I've never thought about formalizing it.
• antognini 11y ago
  Tensors can be somewhat intimidating, but it definitely would not take a year to understand tensors. Two references I found particularly helpful: There's a new textbook on classical physics by Kip Thorne and Roger Blandford. It hasn't been published yet, but the lecture notes on which it's based are online here: http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/ http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/ The first chapter presents a pretty simple introduction to tensors. The other reference I've found helpful is Chapter 31 of Volume II of the Feynman Lectures on Physics. Feynman (as usual) gives a great introduction to the concept. To echo krastanov, it is the case that all tensors can be represented as multi-dimensional arrays, but not all multi-dimensional arrays are tensors. At least in physics, a tensor is ultimately a thing, or at least some description of a physical thing. It therefore has some existence independent of whatever coordinate system you use and therefore, if you change your coordinate system, the values of the tensor have to change in certain ways. This means that your tensor cannot, in general be some arbitrary multidimensional array. There's a bit of confusion about nomenclature since in fields outside physics and math a tensor doesn't need to represent some physical thing so the special transformation properties of a tensor are not so important and it's often just used as a shorthand term for "multidimensional array."
  • mhartl 11y ago
    There's a new textbook on classical physics by Kip Thorne and Roger Blandford. It hasn't been published yet, but the lecture notes on which it's based are online here Heh, it's not exactly new, and I'll believe it's been published when I have it in my hands. That book had already been in development for years—and was due out any day now—back when I used it in Physics 136 (Blandford himself taught one of the terms). That was 1997. The material is great, though, including the coverage of tensors, which served me well the next year in General Relativity (Thorne's last time teaching it). If you're looking for a solid intro to tensor algebra & analysis, I definitely recommend it.
    • bluemanshoe2 11y ago
      It has been a long time coming, but it seems serious now. You can preorder it on amazon: http://www.amazon.com/Modern-Classical-Physics-Elasticity-Statistical-ebook/dp/B00QOBJD7U/ http://www.amazon.com/Modern-Classical-Physics-Elasticity-St... Publication Date: May 31, 2016
• orbifold 11y ago
  If you know the graphical birdtracks/Penrose diagram/Feynman diagram/String diagram notation, then tensors aren't hard at all. Formally what you need to understand is the "tensor product": Given two vector spaces V,W over some field k (think n-tuples of real numbers), their tensor product is characterized by an universal property: Any bilinear map from the Cartesian product V x W to some other vector space U induces a unique linear map from the tensor product V o W to U. You can then proceed to show that the tensor product of vector spaces is associative (V o W) o U == V o (W o U) and symmetric V o W == W o V and the field k is its unit k o V == V == V o k. Interestingly if you are now given two k-linear maps f : V -> V' and g: W -> W' then you can get a map f o g : V o W -> V' o W'. And in the string diagram notation, you write that as V | W | | | f g | | V'| W'| Now an arbitrary tensor is just a linear map a : V1 o .. o Vn -> W1 o .. o Wm. V1 | ... | Vn | ... | ~~~~~~~~~ | a | ~~~~~~~~~ W1| ... | Wm Given a k-vector space V, you can also consider the dual V* vector space of k-linear maps f : V -> k to k. In the diagram notation, you need to then introduce arrows and evaluation and co-evaluation maps V o V* -> k and k -> V o V*, which can be visualized as cups and caps. In the standard notation the distinction between vector spaces and dual vector spaces manifests itself in covariant and contravariant indices of tensors. Differential geometry only enters the picture if you want to study "vector space information", for example the vector space of all possible directions at a point also known as the tangent bundle, attached to points of some geometric space. The simplest manifestation of that concept are vector bundles, since forming the tensor product is an example of a "smooth functor" you can also define a tensor product of vector bundles. What physicists call tensors are sections, that is a "smooth" assignment of a vector in that tensor product at every point, of such vector bundles (typically the tangent bundle/cotangent bundle or some vector bundle associated to a principal bundle of "symmetries")
• rndn 11y ago
  See also "An Introduction to Tensors for Students of Physics and Engineering (2002) [pdf]": https://news.ycombinator.com/item?id=8651645 https://news.ycombinator.com/item?id=8651645
• jheriko 11y ago
  its because they aren't talking about what the majority mean when they use the word "tensor". try looking into "psuedotensor". that is the established word for this for nearly a century. you may have better luck... although probably not because the main use of those historically has been in combination with tensors and tensor fields in the geometric setting and physics.
• obastani 11y ago
  I see a lot of answers describing what tensors are, but none really describe why they're important. To understand this, let's go back to some first-year Calculus. If we have a function f, we can approximate f as f(x+dx) = f(x) + f'(x)dx + O(dx^2) This should look familiar: taking a = f(x) and b = f'(x), this is just the line a + b.dx! In other words, Calculus is just a way transforming questions about (differentiable) functions into questions about lines. So this is all well and good, but what if x is a vector? Or f(x) is a vector? Or both? Well, now we can approximate f(x+dx) = f(x) + Df(x).dx + O(|dx|^2) Here, Df(x) is nothing other than the matrix [(df_i/dx_j)(x)]. In other words, we still get a linear approximation. But now suppose we get greedy and want to take higher order derivatives: what is the derivative of Df(x)? It's a tensor! There are a few ways to think about this (which gives rise to the different interpretations of a tensor): 1. It's just df_i/(dx_jdx_k). This is a multi-dimensional array. 2. In 1D, we have f(x+dx) = f(x) + f'(x)dx + (1/2)f''(x)dx^2 + O(dx^3) In higher dimensions, we get f(x+dx) = f(x) + Df(x)*dx + dx'.D^2f(x).dx + O(|dx|^3) Here, D^2f(x) is a multi-linear map: it takes in two copies of the vector dx, and returns a new vector. More generalizations exist for various reasons, but I think this intuition captures why tensors show up so often.
• formulaT 11y ago
  In this context, tensors are just multidimensional arrays. The easiest way to understand them is by analogy with matrices. Matrices are 2d arrays, and matrix decomposition uses methods inspired by geometry, to analyze data in matrix form. Even though the matrix in question might have no intrinsic geometric meaning (e.g. a matrix of which individual likes which beer), the geometric methods still give useful result. Similarly, tensor decompositions are inspired by geometry, but are also a simple way to get information from a 3d array. For tensors, the most basic tensor decomposition, where a tensor a_ijk is approximated by sum_l b_li * c_lj * d_lk. tl;dr don't read too much into the geometric side of tensors. Data science applications don't have a geometric interpretation, but the techniques can still work.
• deleted 11y ago
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• animakumar 11y ago
  In the podcast, I mentioned tensor as a multidimensional array to simplify the concept and introduce it to a wider audience. Our algorithms certainly treat tensor as a multilinear map and we design efficient decomposition methods. Most importantly, we do not form these tensors, meaning we do not instantiate them as multidimensional arrays. Instead, we implicitly manipulate the data to obtain decompositions of tensors that model higher order relationships in data.