# Tag Info

65

If we're working with three-dimensional vectors, a matrix is a $3\times 3$ array of 9 numbers. If I'm understanding your question right, you're asking whether there is something like a $3\times 3\times 3$ array of 27 numbers with interesting properties. Yes, there is such a thing; it is called a tensor. Tensors are a generalization of both vectors and ...

20

The "product" of vector spaces is more properly thought of as a direct sum, since the dimension increases linearly according to addition, $\dim(V\oplus W)=\dim(V)+\dim(W)$. Tensor products are much bigger in size than sums, since we have $\dim(V\otimes W)=\dim(V)\times\dim(W)$. In fact, in analogy to elementary arithmetic, we have distributivity $(A\oplus ... 17 In addition to the canonical answer involving tensors and multilinear algebra, there is also an approach where the notion of determinant as a solution condition for a system of equations is generalized to some higher dimensional situations. The basic reference for this program (or one form of it) is the book by Gelfand, Kapranov and Zelevinsky of which the ... 14 Matrices are like tables, with elements$A_{m,n}$, with operations of addition and multiplication$(A+B)_{mn} = A_{mn}+B_{mn}$and$(A \cdot B)_{mn} = \sum_k A_{mk} B_{kn}$. Cubic matrices have three indexes$A_{mnk}$, and$(A+B)_{mnk} = A_{mnk}+B_{mnk}$and$(A \cdot B)_{m n k} = \sum_{\ell} A_{m n \ell} B_{m \ell k} C_{\ell n k}$. See ... 13 The first definition comes from the philosophy that students are bad at understanding abstract definitions and would prefer to see the tensor product defined as a space of functions of some kind. This is the reason that some books define the tensor product of$V$and$W$to simply be the space of bilinear functions$V \times W \to k$($k$the underlying ... 11 Let's first set some terminology. Let$V$be an$n$-dimensional real vector space, and let$V^*$denote its dual space. We let$V^k = V \times \cdots \times V$($k$times). A tensor of type$(r,s)$on$V$is a multilinear map$T\colon V^r \times (V^*)^s \to \mathbb{R}$. A covariant$k$-tensor on$V$is a multilinear map$T\colon V^k \to \mathbb{R}$. In ... 11 The answer is yes. There are many places in mathematics where it would be useful to 'store' numbers/whatever in a 3-dimensional grid. That is not the problem. It's using them in a context and defining the right operations that make sense so you can combine things and do some abstract algebra. For a specific example, start with something concrete. ... 11 The exposition you're quoting may be somewhat confusing if you're used to how tensor products are usually defined in a general setting (see e.g. the Wikipedia article). Usually, the tensor product is defined as a new vector space, either through a universal property or by explicit construction using equivalence classes. In your case however, it's being ... 10$\newcommand\P{\mathbb{P}}$Let$V$and$W$be complex vector spaces, and let$\P(V)$,$\P(W)$and$\P(V\otimes W)$be the projective spaces attached to$V$,$W$and$V\otimes W$, respectively. If$v\in V$is non-zero, I'll write$[v]$the point of$\P(V)$corresponding to it; it is the equivalence class of$v$in$V\setminus0$for an the equivalence relation ... 10 While at the vector space level, the pairing might seem slightly forced, we can derive it naturally by adding structure. Given a vector space$V$, we have a graded commutative ring$\bigwedge V = \bigoplus_i \bigwedge^i V$. Given$\phi\in V^*$, it naturally extends to a (graded) derivation$d_{\phi}$of degree$-1$on$\bigwedge V$. Since$d_{\phi}^2=0$... 10 If$M$is a module over$R$, any endomorphism$\phi:M\to M$induces an endomorphism$\Lambda^r \phi:\Lambda^rM\to \Lambda^rM$. If$M$is free with basis$e_1,...,e_n$, then$\phi $has a matrix$A=(a_{ij})$in this basis. The module$ \Lambda^rM$is also free, with basis$(e_H)_{H\in\mathcal H}$where$\mathcal H$is the set of strictly increasing sequences ... 10 I'm going to go way out on a limb and instead of answering the questions actually posed, I'll propose a way to think about..... um OK, here it is: what's the difference between an ordered pair of vectors and a tensor product of two vectors? It is this: If you multiply one of the two vectors by$c$and the other by$1/c$, then you've got a different ordered ... 9 Yes, that's true. Let$f_i : V_i \to W_i$be two linear maps. Since$\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$embeds into$W_1 \otimes W_2$, we may assume that$f_1,f_2$are surjective. But then they are split, so that we can assume that$V_i = W_i \oplus U_i$and that$f_i$equals the projection$V_i \to W_i$, with kernel$U_i$. Then$V_1 \otimes V_2 = ...

9

Tensor products turn multilinear algebra into linear algebra. That's the point (or at least one point). They let you treat different kinds of base extension (e.g., viewing a real matrix as a complex matrix, making a polynomial in ${\mathbf Z}[X]$ into a polynomial in $({\mathbf Z}/m{\mathbf Z})[X]$, turning a representation of a subgroup $H$ into a ...

8

Any approach to multivariable calculus necessitates some modicum of multilinear algebra ( for instance because the change of variables formula for integrals uses determinants), and the cleaner the interface between calculus and algebra, the better. The language of differential forms is a clean interface between the two, and this language generalizes from ...

8

Your first and biggest misconception is that you seem to be mixing up real and complex dimensions. A Riemann surface is a $2$-dimensional real manifold, but as a complex manifold it is $1$-dimensional. Hence you have just one local complex coordinate on $M$, namely $z$. The local coordinates are not $(z, \bar{z})$. There is more structure on the tangent ...

8

This business about working over a commutative ring $R$ is a red herring. Ultimately this is a collection of $n$ polynomial identities in $n^2$ variables $x_{ij}$ over the integers; that is, it suffices to prove this identity over $\mathbb{Z}[x_{ij}]$ as an equality of integer polynomials. But two integer polynomials are equal abstractly if and only if ...

8

Abstract indices is a good alternative to work with, just don't be afraid of using different ranges of symbols dedicated to each bundle you deal with. For instance, I prefer to use the letters $a$, $b$, $c$ etc (the initial segment of the Latin alphabet) as the labels for the tangent and cotangent bundles (depending on the position), and juxtapositions of ...

8

For simplicity, I will only talk about the construction of the tensor product of two $\mathbb{K}$-spaces, $L_1$ and $L_2$. The generalization to the tensor product of any finite number of spaces follows easily. The intuition is this: Whatever the tensor product $L_1 \otimes L_2$ turns out to be, I want it to be generated by elements of the form $l_1 \otimes ... 7 The real trick is to do algebra with modules, rather than in modules. $$M_A \otimes_A N \cong (A \otimes_R M) \otimes_A N \cong (M \otimes_R A) \otimes_A N \cong M \otimes_R (A \otimes_A N) \cong M \otimes_R N$$ That said, to aid with working through your ideas, keep in mind that, usually, when rewriting the product of$a$,$m$, and$n$, it still has ... 7 Where did the meaning of "polarization", in this context, come from? Weyl uses it in his book The classical groups (see pp. 5 and 6 on Google books) but I don't know if this is the first place it appeared. A few things I've managed to find ... The term polarization in this context did not originate with Weyl (1939). The book Hilbert's Invariant ... 7 Some googling brought up this, Hudson's book "Kummer's quartic surface", containing the following argument. Given$A = \begin{pmatrix} B & C\\ D & E\end{pmatrix}$, orthgonality gives us that$A^{-1} = A^t = \begin{pmatrix} B^t & D^t\\C^t & E^t\end{pmatrix}$. The block matrix multiplication gives $$I = A^t A = \begin{pmatrix} B^tB + D^tD ... 7 Let me answer some questions you could be asking, which are more basic than the questions you're actually asking. Let V be a 3-dimensional real vector space. The exterior product gives a canonical pairing$$V \times \Lambda^2(V) \to \Lambda^3(V).$$Now, \Lambda^3(V) is 1-dimensional, but it doesn't come with a canonical isomorphism to \mathbb{R}. ... 6 We induct on n - k. Suppose there exists a nontrivial linear dependence among the pure tensors you list in \Lambda^k(V). Then not all of the tensors share the same components e_i, so there is some i such that some tensor does not contain e_i (and some other tensor does). Take the exterior product with e_i; this gives a linear dependence among a ... 6 You can talk about "isomorphic" maps without problems: that means isomorphism in a category of maps. But we can avoid talking about categories of maps and safely delete the word "category" in this case: you have just two maps with the same source$$ X \stackrel{b}{\longleftarrow} U \times V \stackrel{g}{\longrightarrow} U \otimes V \ . $$For these two ... 6 I approach the question from a computer programmer's point of view. Most programming languages have array constructs. An array is an ordered sequence of elements, where each element might be a number (integer or real) or something else. You can also have multidimensional arrays; depending on the programming language, they might simply be arrays of arrays, ... 6 Suppose that D:F^n\rightarrow F is n-linear. In other words,$$D(ca_1,\ldots,a_n)=cD(a_1,\ldots,a_n)\vdotsD(a_1,\ldots,ca_n)=cD(a_1,\ldots,a_n)$$for all a_1,\ldots,a_n,c\in F, and$$D(a_1+b,a_2,\ldots,a_n)=D(a_1,a_2,\ldots,a_n)+D(b,a_2,\ldots,a_n)\vdotsD(a_1,\ldots,a_{n-1},a_n+b)=D(a_1,\ldots,a_{n-1},a_n)+D(a_1,\ldots,a_{n-1},b)$$... 6 You're right: this is a notation failure (on the part of many differential geometry texts). The correct notation should be$\bigwedge^k (V^*)$, as you say. However, confusingly many texts write$\bigwedge^k(V)$to mean$\bigwedge^k(V^*)$. (edit by EA, 5:54. fixed a typo) 6$\newcommand{\tr}{\operatorname{tr}}$First, let us give an alternative description of$\tr {\bigwedge}^m T$. Let$e_1,e_2,\dots,e_n$be a fixed basis for$V$, and for$J \subset \{1,2,\dots,n\}$with$|J| = m$, let$T_J$denote the$m \times m$matrix acquired by keeping the entries$T_{i,j}$with$i,j \in J$. Alternatively, if$P_J\$ is the projection ...

Only top voted, non community-wiki answers of a minimum length are eligible