123acsiac Posted February 28, 2011 Share Posted February 28, 2011 Hi all, My question is: Is there a good source that defines "tensor rank" for me? Yes, I have seen several already, but perhaps I am mostly interested in a definition that is more mathematical (rigid)...For example, how do we define the rank of a multilinear form? If I have f: V x V x V...x V --------> k, what should the rank of this map be? Thank you Link to comment Share on other sites More sharing options...

ydoaPs Posted February 28, 2011 Share Posted February 28, 2011 NASA has a good pdf on tensors. Link to comment Share on other sites More sharing options...

DrRocket Posted February 28, 2011 Share Posted February 28, 2011 (edited) Hi all, My question is: Is there a good source that defines "tensor rank" for me? Yes, I have seen several already, but perhaps I am mostly interested in a definition that is more mathematical (rigid)...For example, how do we define the rank of a multilinear form? If I have f: V x V x V...x V --------> k, what should the rank of this map be? Thank you k In physics a "tensor" is usually really a tensor field on some manifold. A covariant tensor is a tensor field, a field of multilinear forms, on the tangent bundle. A contravariant tensor field acts on the cotangent bundle. A tensor of order, or type, (q,p) or (q+p) acts on T x T x ... x T x T* x T* x ... x T* with q copies of T and p copies of T*. People who fool around with this stuff regularly do this without thinking. I have to stop and remember what the notation means. If you haven't become confused by this yet, you will be sometime. Edited February 28, 2011 by DrRocket Link to comment Share on other sites More sharing options...

ajb Posted February 28, 2011 Share Posted February 28, 2011 A covariant tensor is a tensor field, a field of multilinear forms, on the tangent bundle. A contravariant tensor field acts on the cotangent bundle. A tensor of order, or type, (q,p) or (q+p) acts on T x T x ... x T x T* x T* x ... x T* with q copies of T and p copies of T*. Using spaces, rather than natural bundles over manifolds we have [math]\underbrace{V \times V \times \cdots \times V}_{q} \times \underbrace{V^{*} \times V^{*} \times \cdots \times V^{*}}_{p} \rightarrow K[/math], where [math]K[/math] is our ground field, probably [math]R[/math]. Over a smooth manifold tensors, or rather tensor fields are a kind of geometric object, which are understood as sections of natural bundles. Thus, as DrRocket states, tensors can be thought of as sections of vector bundles of the form [math]TM \times TM \times \cdots \times T^{*}M \times \cdots \times T^{*}M \rightarrow M[/math] and various symmetrisations and antisymmetrisation. A section is a map from the manifold [math]M[/math] to the bundle. For instance, a vector field can be understood in several ways. Either as a derivation of the algebra of smooth functions over a differentiable manifold, so [math] X = X^{A}(x) \frac{\partial}{\partial x^{A}}[/math], in some local coordinates or as a section of the tangent bundle [math]X = X^{A}e_{A}[/math], where [math]e_{A}[/math] is a basis of sections. Link to comment Share on other sites More sharing options...

Xerxes Posted February 28, 2011 Share Posted February 28, 2011 Ugh! This is a subject I thought I knew fairly well. Obviously I was mistaken, as I find the following very confusing Using spaces, rather than natural bundles over manifolds we have [math]\underbrace{V \times V \times \cdots \times V}_{q} \times \underbrace{V^{*} \times V^{*} \times \cdots \times V^{*}}_{p} \rightarrow K[/math], where [math]K[/math] is our ground field, probably [math]R[/math]. It was to my understanding that the above is the domain of some multilinear form aka tensor. Simplifying, let's say, [math]V \otimes V^{*} :V^{*} \times V \to \mathbb{R}[/math]. But OK, in your example the rank of your tensor is simply p + q, and mine it is 1 + 1 =2 , tensors {meaning tensor fields, I assume} can be thought of as sections of vector bundles of the form [math]TM \times TM \times \cdots \times T^{*}M \times \cdots \times T^{*}M \rightarrow M[/math] Tensors do not need to thought of as fields, but no matter. You are, of course, entitled to use whatever notation you choose, but a tangent space at the arbitrary point [math]p \in M[/math] is generally referred to as [math]T_pM[/math], and the tangent bundle, the set theoretic (disjoint) union of all such tangent spaces as [math]TM[/math]. It seems (at least to me) that to any manifold one can associate only a single tangent bundle. Is this wrong? Whatever, I cannot make sense of the Cartesian product of bundles, assuming you are using standard notation. A section is a map from the manifold [math]M[/math] to the bundle. Again, I am confused by this. Taking the intuitive view that a vector field on [math]M[/math] assigns (smoothly) to every point [math]p \in M[/math] a single element from its associated tangent space [math]T_pM[/math], this seems to imply that a field is a section that is a map from the bundle to the base manifold, not the other way around. Maybe it's me that's the other around Link to comment Share on other sites More sharing options...

ajb Posted February 28, 2011 Share Posted February 28, 2011 (edited) It was to my understanding that the above is the domain of some multilinear form aka tensor. Simplifying, let's say, [math]V \otimes V^{*} :V^{*} \times V \to \mathbb{R}[/math]. But OK, in your example the rank of your tensor is simply p + q, and mine it is 1 + 1 =2 [math]V^{*} \times V \to \mathbb{R}[/math] has tensor rank two. It has "one lot of the space and one lot of the dual". Tensors do not need to thought of as fields, but no matter. You are, of course, entitled to use whatever notation you choose, but a tangent space at the arbitrary point [math]p \in M[/math] is generally referred to as [math]T_pM[/math], and the tangent bundle, the set theoretic (disjoint) union of all such tangent spaces as [math]TM[/math]. By a tensor, it depends on exactly what you mean. You can of course consider a "tensor" (or other geometric object) at a point, this what can be meant by tensor. A smooth assignment of a tensor to every point on a smooth manifold is a tensor field. Often by tensor we really mean tensor field, but you are right that one should be careful with this distinction. It seems (at least to me) that to any manifold one can associate only a single tangent bundle. Is this wrong? Whatever, I cannot make sense of the Cartesian product of bundles, assuming you are using standard notation. You do it all fibrewise. Again, I am confused by this. Taking the intuitive view that a vector field on [math]M[/math] assigns (smoothly) to every point [math]p \in M[/math] a single element from its associated tangent space [math]T_pM[/math], this seems to imply that a field is a section that is a map from the bundle to the base manifold, not the other way around. Maybe it's me that's the other around The projection is a map from the tangent bundle to the base manifold. The little rhyme to remember is "sections undo projections", in a sense (you can make this precise) sections are the "inverse" of the projection. EDIT: Also elements of the vector space [math]V \times V \cdots \times V \times V^{*} \times V^{*}\cdots V^{*}[/math] for a vector space [math]V[/math] are known as tensors. Edited February 28, 2011 by ajb Link to comment Share on other sites More sharing options...

Xerxes Posted March 1, 2011 Share Posted March 1, 2011 (edited) Yikes! this seems to imply that a field is a section that is a map from the bundle to the base manifold, not the other way around. So moron wants the mapping that assigns a vector to each point [math]p \in M[/math] a single element [math]v \in T_pM[/math] to have the point [math]p \in M[/math] as an image point? Jeez, that's bonkers! Also elements of the vector space [math]V \times V \cdots \times V \times V^{*} \times V^{*}\cdots V^{*}[/math] for a vector space [math]V[/math] are known as tensors. Confidence shaken, I ask in trepidation: What does this mean? As far as I am aware, what you wrote is the domain of the tensor aka multilinear map [math]V^{*} \otimes V^{*} \cdots \otimes V^{*} \otimes V \otimes V\cdots V[/math] whose codomain is Real. How can this be a tensor? Surely it is nothing more (or less) than the Cartesian product of vector spaces. Does this make it a tensor? Or am I wrong again? Edited March 1, 2011 by Xerxes Link to comment Share on other sites More sharing options...

ajb Posted March 1, 2011 Share Posted March 1, 2011 (edited) Confidence shaken, I ask in trepidation: What does this mean? As far as I am aware, what you wrote is the domain of the tensor aka multilinear map [math]V^{*} \otimes V^{*} \cdots \otimes V^{*} \otimes V \otimes V\cdots V[/math] whose codomain is Real. How can this be a tensor? Surely it is nothing more (or less) than the Cartesian product of vector spaces. Does this make it a tensor? Some people call an element of the tensor product of vector spaces and their dual a tensor. (I probably should have made the tensor product a lot clearer) so [math]T \in V^{*} \otimes V^{*} \cdots \otimes V^{*} \otimes V \otimes V\cdots V[/math] is called a tensor. Some people call the multilinear map [math]V^{*}\times V^{*} \times \cdots V \times V \cdots V \rightarrow R[/math] a tensor. Then you have tensor fields on smooth manifolds that have several ways of defining them. These too are often just called tensors. So, I would say that exactly what you mean by a tensor will depend on the context. Edited March 1, 2011 by ajb Link to comment Share on other sites More sharing options...

DrRocket Posted March 1, 2011 Share Posted March 1, 2011 Some people call an element of the tensor product of vector spaces and their dual a tensor. (I probably should have made the tensor product a lot clearer) so [math]T \in V^{*} \otimes V^{*} \cdots \otimes V^{*} \otimes V \otimes V\cdots V[/math] is called a tensor. Some people call the multilinear map [math]V^{*}\times V^{*} \times \cdots V \times V \cdots V \rightarrow R[/math] a tensor. Then you have tensor fields on smooth manifolds that have several ways of defining them. These too are often just called tensors. So, I would say that exactly what you mean by a tensor will depend on the context. One can always revert to the physicist's definition: A tensor is a symbol having four corners, the two on the right being particularly appealing, to which one may attach an unlimited number of indices. Link to comment Share on other sites More sharing options...

Xerxes Posted March 1, 2011 Share Posted March 1, 2011 Some people call an element of the tensor product of vector spaces and their dual a tensor. (I probably should have made the tensor product a lot clearer) so [math]T \in V^{*} \otimes V^{*} \cdots \otimes V^{*} \otimes V \otimes V\cdots V[/math] is called a tensor. Yes, that is the definition I would use. I have seen another (see below) Some people call the multilinear map [math]V^{*}\times V^{*} \times \cdots V \times V \cdots V \rightarrow R[/math] a tensor. Which does not differ significantly from the above, except it has a slight category-theoretic feel to it: the arrow is the tensor! So, I would say that exactly what you mean by a tensor will depend on the context. OK, in applications, particularly in differential geometry, it is common to define a tensor in terms of how it transforms under a coordinate change. This is not, I believe, very helpful as a definition, though I grant the transformation rules are essential to understanding tensors in this context. PS I do not understand DrRocket's post at all Link to comment Share on other sites More sharing options...

DrRocket Posted March 1, 2011 Share Posted March 1, 2011 Yes, that is the definition I would use. I have seen another (see below) Which does not differ significantly from the above, except it has a slight category-theoretic feel to it: the arrow is the tensor! Let k be a commutative ring and let[math]E_1,E_2,...,E_n[/math] be k-modules. Consider the class L of all multilinear functions [math]f:E_1 \times E_2 \times ... \times E_n \rightarrow F[/math] where [math]F[/math] varies with [math]f[/math]. The elements may be viewed as the objects of a category. If [math]f:E_1 \times E_2 \times ... \times E_n \rightarrow F[/math] and [math]g:E_1 \times E_2 \times ... \times E_n \rightarrow G[/math] are two objects in L, a morphism [math]f \rightarrow g[/math] is a homomorphism [math]h:F \rightarrow G[/math] such that [math] h \circ f = g [/math] The tensor product of [math]E_1,E_2,...,E_n[/math] denoted [math]E_1 \otimes E_2 \otimes ... \otimes E_n[/math] is a universal repelling object in this category. Link to comment Share on other sites More sharing options...

ajb Posted March 2, 2011 Share Posted March 2, 2011 OK, in applications, particularly in differential geometry, it is common to define a tensor in terms of how it transforms under a coordinate change. This is not, I believe, very helpful as a definition, though I grant the transformation rules are essential to understanding tensors in this context. Right, so a tensor is defined as a special type of geometric object. Rather than repeating myself have a look here. I think that defining tensors as a geometric object is far from elegant, but it is very clear what you mean and very practical for most applications. The modern definition of a geometric object is in terms of natural bundles. So, a natural bundle is a functor from the category of smooth manifolds (or supermanifolds) to the category of smooth bundles such that local differomorphisms (changes of coordinate) become bundle automorphisms. In short, these are the bundles that "come for free" once you a smooth structure on a manifold. Examples include the tangent and cotangent bundle. A geometric object is a section of a natural bundle. Tensors are sections of natural vector bundles. Link to comment Share on other sites More sharing options...

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