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This document tells an important lie: it says (e.g., at the top of page 8) that every rank-2 tensor is an outer product of two vectors. This is flatly untrue an

by gjm11 12y ago

This document tells an important lie: it says (e.g., at the top of page 8) that every rank-2 tensor is an outer product of two vectors. This is flatly untrue and is an easy misconception for a student to acquire even if the thing they're learning from doesn't explicitly say it. I skimmed through looking to see if at some point the author says "oh, by the way, I told you a lie earlier", but it doesn't look like he does. (If anyone reading this has difficulty believing that there are more rank-2 tensors than outer products ("dyads") of two vectors, note that in 3 dimensions you can specify two vectors by giving 6 numbers, but it takes 9 to specify a rank-2 tensor because it can be represented by an arbitrary 3x3 matrix.)

evanb 12y ago
Indeed, this is one of the hallmarks of entanglement in quantum mechanics. Suppose you have two qubits, one of which can be in the states |0> and |1> and the other which can be in the states |a> and |b> (an unconventional choice, but clearer when there's no TeX). If |psi> = some superposition of |0> and |1> and |chi> = some superposition of |a> and |b> then there is a class of states |psi> (tensor product) |chi>. These states factorize, and thus the two qubits are not entangled. However, there are states like |bell> = |0> (tensor) |a> + |1> (tensor) |b> which do not factorize. We call these states entangled.
lipak45 12y ago
Items 1-3 in the "Specific Statements for Tensors per se" (pqge 6) seems to allude to your point. "1. All scalars are not tensors, although all tensors of rank 0 are scalars (see below). 2. All vectors are not tensors, although all tensors of rank 1 are vectors (see below). 3. All dyads or matrices are not tensors, although all tensors of rank 2 are dyads or matrices."
- gjm11 12y ago
  It's not perfectly clear whether "dyads or matrices" means "dyads, which we also call matrices" (wrong!) or "either dyads or matrices" (right, kinda!), which is why I picked on a slightly later bit of the document to support my accusation. It's also not perfectly clear what the author really means by saying that not all matrices "are" tensors but all tensors "are" matrices; this is part of what he says is summarizing previous remarks but I don't see any previous remarks addressing the point. The usual answer is to say that a tensor is a matrix that "transforms correctly"; in something closer to puremathematicianese, what this really means is that we're looking not exactly at matrices but at functions from coordinate systems to matrices (i.e., you plug in the coordinate system you want to use, and out come the components of the thing-that-might-be-a-tensor as a 3x3 array), and that what you've got is a tensor of a particular kind if and only if the way in which these numbers change as the coordinate system does is right. But at this point in the document he hasn't said anything remotely like that. He does eventually mention it but not until the very end, where he seems to think he's getting it as a consequence of some other definition -- but unless that definition is the (definitely wrong) one that says the only tensors are tensor products of vectors, it's not clear what it is. (It is in fact possible to define tensors in another way that's neater and coordinate-free, and derive the transformation rules as theorems. One way to do this is closely related to what this document does, but instead of (wrongly) saying that all tensors are tensor products of vectors ("dyads", etc.) what you (rightly) say is that all tensors are linear combinations of tensor products of vectors. Er, and their duals. And you actually have to say what all that means, and it turns out to be a rather elegant bit of theory. Perhaps that's what this document is gesturing towards, but it really doesn't do more than gesture.)
  - jules 12y ago
    I have exactly this experience whenever I read something by a physicist! It is another universe where everything seems vague, but they somehow manage to get the correct answer anyway, and usually very quickly too. > and it turns out to be a rather elegant bit of theory Do you have a recommendation on what to read for that?
thomasahle 12y ago
> If anyone reading this has difficulty believing that there are more rank-2 tensors than outer products Small nitpick: Unless you have tensors over a finite universe, there aren't really "more" of the large ones that the small ones.. So we have to argue in some other way. Anybody's got a counter example? (matrix that is probably not a dyad?)
- gjm11 12y ago
  Yeah, there really are more, provided you define "more" right. They form a topological space, and the rank-2 tensors are a 9-dimensional space versus a 6-dimensional space for the outer products. Alternatively, any matrix whose determinant isn't zero is a non-dyad.
  - sampo 12y ago
    But the cardinality of the set of all points in a 9-dimensional space is not larger than the cardinality of the set of points in a 6-dimensional space, or even between 0 and 1 in the 1-dimensional real line. And the set theoretic cardinality is the normal way to define what "more" means, or to compare sets of an infinite amount of objects. How would you define "more", to get e.g. R² have "more" points than R?
    - wetmore 12y ago
      You can say that R^2 has "more" points than R by inclusion: there is a natural injection of R into R^2 as a proper subspace. Think of the odd numbers in the integers, for example. "More" is a fairly loose concept in that there is no strict definition. You can use it to refer to cardinality, or some other sense of order.
    - gjm11 12y ago
      There are lots of ways to define "more" (or "bigger" or whatever term one chooses to use). Which one is most appropriate depends on what kind of things you're working with. Working with vector spaces over a given field: "bigger" means "higher dimension". Working with manifolds: "bigger" means "higher dimension". Working with sets: "bigger" usually means "larger cardinality" but could sometimes mean "proper superset". In general, you can take "X is bigger than Y" to mean "Y is isomorphic to some substructure of X, but not vice versa". If "isomorphic" means "in bijection with" and "substructure" means "subset" then this is cardinality. If they mean "isomorphic as vector spaces over F" and "vector subspace" then it's dimension. If they mean "homeomorphic" and "topological subspace" then actually it isn't true that lower-dimensional manifolds are always "smaller than" higher-dimensional ones (so maybe we actually want a different definition) but in this particular case the rank-1 tensors are a lower-dimensional proper subspace of all the tensors so the definition gives the "right" answer.
hyperliner 12y ago
Well, it's really not necessarily a "lie" as much as it is a "mistake." Not everything that is "flatly untrue" is a "lie" as there is no apparent intent to deceive. Basically, when you say that "this document tells an important lie" you are likely lying. At least you are accusing somebody of being a liar.
sampo 12y ago
> This document tells an important lie: it says (e.g., at the top of page 8) that every rank-2 tensor is an outer product of two vectors. Then again, that whole section is in the context of material stresses and permeability, which are described as symmetrical tensors, for which the claim is true.

by gjm11 12y ago

evanb 12y ago

lipak45 12y ago

Items 1-3 in the "Specific Statements for Tensors per se" (pqge 6) seems to allude to your point. "1. All scalars are not tensors, although all tensors of rank 0 are scalars (see below). 2. All vectors are not tensors, although all tensors of rank 1 are vectors (see below). 3. All dyads or matrices are not tensors, although all tensors of rank 2 are dyads or matrices."

gjm11 12y ago
It's not perfectly clear whether "dyads or matrices" means "dyads, which we also call matrices" (wrong!) or "either dyads or matrices" (right, kinda!), which is why I picked on a slightly later bit of the document to support my accusation. It's also not perfectly clear what the author really means by saying that not all matrices "are" tensors but all tensors "are" matrices; this is part of what he says is summarizing previous remarks but I don't see any previous remarks addressing the point. The usual answer is to say that a tensor is a matrix that "transforms correctly"; in something closer to puremathematicianese, what this really means is that we're looking not exactly at matrices but at functions from coordinate systems to matrices (i.e., you plug in the coordinate system you want to use, and out come the components of the thing-that-might-be-a-tensor as a 3x3 array), and that what you've got is a tensor of a particular kind if and only if the way in which these numbers change as the coordinate system does is right. But at this point in the document he hasn't said anything remotely like that. He does eventually mention it but not until the very end, where he seems to think he's getting it as a consequence of some other definition -- but unless that definition is the (definitely wrong) one that says the only tensors are tensor products of vectors, it's not clear what it is. (It is in fact possible to define tensors in another way that's neater and coordinate-free, and derive the transformation rules as theorems. One way to do this is closely related to what this document does, but instead of (wrongly) saying that all tensors are tensor products of vectors ("dyads", etc.) what you (rightly) say is that all tensors are linear combinations of tensor products of vectors. Er, and their duals. And you actually have to say what all that means, and it turns out to be a rather elegant bit of theory. Perhaps that's what this document is gesturing towards, but it really doesn't do more than gesture.)
- jules 12y ago
  I have exactly this experience whenever I read something by a physicist! It is another universe where everything seems vague, but they somehow manage to get the correct answer anyway, and usually very quickly too. > and it turns out to be a rather elegant bit of theory Do you have a recommendation on what to read for that?

thomasahle 12y ago

> If anyone reading this has difficulty believing that there are more rank-2 tensors than outer products Small nitpick: Unless you have tensors over a finite universe, there aren't really "more" of the large ones that the small ones.. So we have to argue in some other way. Anybody's got a counter example? (matrix that is probably not a dyad?)

gjm11 12y ago
Yeah, there really are more, provided you define "more" right. They form a topological space, and the rank-2 tensors are a 9-dimensional space versus a 6-dimensional space for the outer products. Alternatively, any matrix whose determinant isn't zero is a non-dyad.
- sampo 12y ago
  But the cardinality of the set of all points in a 9-dimensional space is not larger than the cardinality of the set of points in a 6-dimensional space, or even between 0 and 1 in the 1-dimensional real line. And the set theoretic cardinality is the normal way to define what "more" means, or to compare sets of an infinite amount of objects. How would you define "more", to get e.g. R² have "more" points than R?
  - wetmore 12y ago
    You can say that R^2 has "more" points than R by inclusion: there is a natural injection of R into R^2 as a proper subspace. Think of the odd numbers in the integers, for example. "More" is a fairly loose concept in that there is no strict definition. You can use it to refer to cardinality, or some other sense of order.
  - gjm11 12y ago
    There are lots of ways to define "more" (or "bigger" or whatever term one chooses to use). Which one is most appropriate depends on what kind of things you're working with. Working with vector spaces over a given field: "bigger" means "higher dimension". Working with manifolds: "bigger" means "higher dimension". Working with sets: "bigger" usually means "larger cardinality" but could sometimes mean "proper superset". In general, you can take "X is bigger than Y" to mean "Y is isomorphic to some substructure of X, but not vice versa". If "isomorphic" means "in bijection with" and "substructure" means "subset" then this is cardinality. If they mean "isomorphic as vector spaces over F" and "vector subspace" then it's dimension. If they mean "homeomorphic" and "topological subspace" then actually it isn't true that lower-dimensional manifolds are always "smaller than" higher-dimensional ones (so maybe we actually want a different definition) but in this particular case the rank-1 tensors are a lower-dimensional proper subspace of all the tensors so the definition gives the "right" answer.

hyperliner 12y ago

Well, it's really not necessarily a "lie" as much as it is a "mistake." Not everything that is "flatly untrue" is a "lie" as there is no apparent intent to deceive. Basically, when you say that "this document tells an important lie" you are likely lying. At least you are accusing somebody of being a liar.

sampo 12y ago

> This document tells an important lie: it says (e.g., at the top of page 8) that every rank-2 tensor is an outer product of two vectors. Then again, that whole section is in the context of material stresses and permeability, which are described as symmetrical tensors, for which the claim is true.