NOTES ON STATISTICS, PROBABILITY and MATHEMATICS
Tensor Product and Multilinear Maps:
In the Wikipedia example of the tensor product vector spaces, included in my OP, as well as in my previous post here, the tensor product is of the form (V\otimes V,) a ((0,2)) tensor, and results in a form akin to ((1)) in the OP:
[A^0 B^0 e_0 \otimes e_0 + A^0 B^1 e_0 \otimes e_1 + \cdots + A^4 B^4 e_4 \otimes e_4]
equivalent to an outer product, as illustrated in this post:
The tensor product of two vectors (v\in V) and (w ...
NOTES ON STATISTICS, PROBABILITY and MATHEMATICS
Tensor Product and Multilinear Maps:
In the Wikipedia example of the tensor product vector spaces, included in my OP, as well as in my previous post here, the tensor product is of the form (V\otimes V,) a ((0,2)) tensor, and results in a form akin to ((1)) in the OP:
[A^0 B^0 e_0 \otimes e_0 + A^0 B^1 e_0 \otimes e_1 + \cdots + A^4 B^4 e_4 \otimes e_4]
equivalent to an outer product, as illustrated in this post:
The tensor product of two vectors (v\in V) and (w \in W), i.e. ((V\otimes W)) is akin to calculating the outer product of two vectors:
[\large v\otimes_o w=\small \begin{bmatrix}-2.3;e_1\+1.9;e_2\-0.5;e_3\end{bmatrix}\begin{bmatrix}0.7;e_1&-0.3;e_2&0.1;e_3\end{bmatrix}= \begin{bmatrix}-1.61;e_1\otimes e_1&+0.69;e_1\otimes e_2&-0.23;e_1\otimes e_3\+1.33;e_2 \otimes e_1&-0.57;e_2 \otimes e_2&+0.19;e_2 \otimes e_3\-0.35;e_3 \otimes e_1&+0.15;e_3 \otimes e_2&-0.05;e_3 \otimes e_3\end{bmatrix}]
This is equivalent to the tensor product space (V^*\otimes V^*) (the set of all tensor ((2,0))) on the slide in the OP. The presenter is tensor-multiplying two co-vectors in the vector basis of (V^*), without coefficients, yielding the (16) pairs of basis vectors of (V^*\otimes V^*): [{e^0\otimes e^0, ; e^0\otimes e^1, ; e^0\otimes e^3, ;\cdots, e^4\otimes e^4}.]
The key is to distinguish these forms of tensor product of vector spaces from their application to other vectors (or covectors), i.e. when the [\langle\beta_\mu,e^\mu;,;A^\nu,e_\nu \rangle;=\beta_\mu,A^\nu,\langle e^\mu;,;e_\nu\rangle ;=\beta_\mu,A^\nu,\delta^\mu_{;\nu};=\beta_\mu,A^\mu;\in \mathbb R] operations are carried out, yielding a real number - which is what is explained in the video.
These linear mappings (\beta\otimes\gamma:V\times V \to \mathbb R) properly interpreted as ([\beta\otimes\gamma](v,w)=\langle \beta,v\rangle\langle\gamma,w\rangle) (i.e. the tensor (\beta\otimes\gamma) acting on two vectors, (v) and (w)) would correct the ((2)) part of the OP (after Professor Shifrin’s answer) as:
(\begin{align} &(\beta\otimes\gamma)\left(\sum A^\mu e_\mu,\sum B^\nu e_\nu\right)= \[2ex] &=\left [ \beta_0\gamma_0;e^0\otimes e^0+ ; \beta_0\gamma_1;e^0\otimes e^1+ ;\beta_0\gamma_3; e^0\otimes e^3+\cdots+ ;\beta_4\gamma_4; e^4\otimes e^4 \right],\small{\left(\sum A^\mu e_\mu,\sum B^\nu e_\nu\right) } \[2ex] &= \beta_0\gamma_0 A^\mu B^\nu \langle e^0,e_\mu \rangle ; \langle e^0,e_\nu \rangle ; + ; \beta_0\gamma_1 A^\mu B^\nu \langle e^0,e_\mu \rangle ; \langle e^1,e_\nu \rangle +\cdots +\beta_4\gamma_4 A^\mu B^\nu \langle e^4,e_\mu \rangle ; \langle e^4,e_\nu \rangle \[2ex] &=\beta_0\gamma_0 A^\mu B^\nu ; \delta^0_{;\mu}; \delta^0_{;\nu} ; + ; \beta_0\gamma_1 A^\mu B^\nu ; \delta^0_{;\mu}; \delta^1_{;\nu} +\cdots +\beta_4\gamma_4 A^\mu B^\nu ; \delta^4_{;\mu}; \delta^4_{;\nu} \[2ex]&= \sum \beta_\mu\gamma_\nu A^\mu B^\nu \end{align})
indeed a real number, exemplifying the mapping (V\times V \to \mathbb R.) The tensor is defined as
[\begin{align}\beta\otimes \gamma&:= \beta_0\gamma_0, e^0\otimes e^0+\beta_0\gamma_1, e^0\otimes e^1 + \beta_0\gamma_2, e^0\otimes e^2+\cdots+\beta_3\gamma_3, e^3\otimes e^3\[2ex] &=T_{00}, e^0\otimes e^0+T_{01}, e^0\otimes e^1 + T_{02}, e^0\otimes e^2+\cdots+T_{33}, e^3\otimes e^3\[2ex] &= T_{\mu\nu},e^\mu\otimes,e^\nu \end{align}]
Tensor Product as the Kronecker product:
As an example, I believe we could illustrate this as follows:
(\beta \in V^*) is (\beta=\color{blue}{\begin{bmatrix}\sqrt{\pi} & \sqrt[3]{\pi} &\sqrt[5]{\pi} \end{bmatrix}}) and (\gamma\in V^*) is (\gamma=\color{red}{\begin{bmatrix}\frac{1}{3} &\frac{1}{5} &\frac{1}{7} \end{bmatrix}}). The ((2,0))-tensor (\beta\otimes \gamma) is the outer product:
[\begin{align}\beta\otimes_o \gamma&= \begin{bmatrix}\color{blue}{\sqrt\pi}\times \color{red}{\frac{1}{3}}\quad e^1\otimes e^1 &\color{blue}{\sqrt\pi}\times\color{red}{\frac{1}{5}}\quad e^1\otimes e^2 &\color{blue}{\sqrt\pi}\times\color{red}{\frac{1}{7}}\quad e^1\otimes e^3\ \color{blue}{\sqrt[3]{\pi}}\times\color{red}{\frac{1}{3}}\quad e^2\otimes e^1 &\color{blue}{\sqrt[3]{\pi}}\times\color{red}{\frac{1}{5}}\quad e^2\otimes e^2 &\color{blue}{\sqrt[3]{\pi}}\times\color{red}{\frac{1}{7}}\quad e^2\otimes e^3 \\color{blue}{\sqrt[5]{\pi}}\times\color{red}{\frac{1}{3}}\quad e^3\otimes e^1 &\color{blue}{\sqrt[5]{\pi}}\times\color{red}{\frac{1}{5}}\quad e^3\otimes e^2 &\color{blue}{\sqrt[5]{\pi}}\times \color{red}{\frac{1}{7}}\quad e^3\otimes e^3\end{bmatrix}\[2ex] &=\begin{bmatrix}\color{red}{\frac{1}{3}}\color{blue}{\sqrt\pi}\quad e^1\otimes e^1&\color{red}{\frac{1}{5}}\color{blue}{\sqrt\pi}\quad e^1\otimes e^2&\color{red}{\frac{1}{7}}\color{blue}{\sqrt\pi}\quad e^1\otimes e^3\\color{red}{\frac{1}{3}}\color{blue}{\sqrt[3]{\pi}}\quad e^2\otimes e^1&\color{red}{\frac{1}{5}}\color{blue}{\sqrt[3]{\pi}}\quad e^2\otimes e^2&\color{red}{\frac{1}{7}}\color{blue}{\sqrt[3]{\pi}}\quad e^2\otimes e^3\\color{red}{\frac{1}{3}}\color{blue}{\sqrt[5]{\pi}}\quad e^3\otimes e^1&\color{red}{\frac{1}{5}}\color{blue}{\sqrt[5]{\pi}}\quad e^3\otimes e^2&\color{red}{\frac{1}{7}} \color{blue}{\sqrt[5]{\pi}}\quad e^3\otimes e^3\end{bmatrix} \end{align}]
Now if we apply this tensor product on the vectors
[v=\color{magenta}{\begin{bmatrix}1\7\5\end{bmatrix}}, w = \color{orange}{\begin{bmatrix}2\0\3\end{bmatrix}}]
[\begin{align} (\beta \otimes \gamma)[v,w]=&\[2ex] & ;\color{blue}{\sqrt\pi}\times \color{red}{\frac{1}{3}} \times \color{magenta} 1 \times \color{orange}2 \quad+\quad \color{blue}{\sqrt\pi}\times\color{red}{\frac{1}{5}} \times \color{magenta}1 \times \color{orange} 0 \quad+\quad \color{blue}{\sqrt\pi}\times,\color{red}{\frac{1}{7}} \times \color{magenta}1 \times \color{orange}3 \ + &;\color{blue}{\sqrt[3]{\pi}}\times\color{red}{\frac{1}{3}} \times \color{magenta}{7} \times \color{orange}2 \quad+\quad \color{blue}{\sqrt[3]{\pi}}\times\color{red}{\frac{1}{5}} \times \color{magenta}{7} \times \color{orange}0 \quad+\quad \color{blue}{\sqrt[3]{\pi}}\times\color{red}{\frac{1}{7}} \times \color{magenta}{7} \times \color{orange}3 \ ;+ &;\color{blue}{\sqrt[5]{\pi}}\times\color{red}{\frac{1}{3}} \times \color{magenta} 5 \times \color{orange}2 \quad+\quad \color{blue}{\sqrt[5]{\pi}}\times\color{red}{\frac{1}{5}} \times \color{magenta} 5 \times \color{orange}0 \quad+\quad \color{blue}{\sqrt[5]{\pi}}\times \color{red}{\frac{1}{7}} \times \color{magenta}5 \times \color{orange}3 \[2ex] =&\ & \color{blue}{\sqrt{\pi}};\times\color{magenta} 1 \quad\left(\color{red}{\frac{1}{3}} \times \color{orange}2 \quad+\quad \color{red}{\frac{1}{5}} \times \color{orange} 0 \quad+\quad \color{red}{\frac{1}{7}} \times \color{orange}3\right) \ + &,\color{blue}{\sqrt[3]\pi} \times \color{magenta}{7}\quad\left(\color{red}{\frac{1}{3}} \times \color{orange}2 \quad+\quad \color{red}{\frac{1}{5}} \times \color{orange}0 \quad+\quad \color{red}{\frac{1}{7}} \times \color{orange}3\right) \ ;+ &,\color{blue}{\sqrt[5]{\pi}}\times \color{magenta} 5\quad\left(\color{red}{\frac{1}{3}} \times \color{orange}2 \quad+\quad \color{red}{\frac{1}{5}} \times \color{orange}0 \quad+\quad \color{red}{\frac{1}{7}} \times \color{orange}3 \right)\[2ex] =&\&\small \left(\color{blue}{\sqrt\pi} \times \color{magenta} 1 \quad+\quad \color{blue}{\sqrt[3]\pi} \times \color{magenta}{7} \quad +\quad \color{blue}{\sqrt[5]\pi} \times \color{magenta}5 \right) \times \left(\color{red}{\frac{1}{3}} \times \color{orange}2 \quad+\quad \color{red}{\frac{1}{5}} \times \color{orange} 0 \quad +\quad \color{red}{\frac{1}{7}} \times \color{orange} 3 \right)\[2ex] =&\[2ex]&\langle \color{blue}\beta,\color{magenta}v \rangle \times \langle \color{red}\gamma,\color{orange}w \rangle\[2ex] =& 20.05487\end{align}]
The elements of the first vector, (v,) multiply separate rows of the outer product (\beta \otimes_o \gamma,) while the elements of the second vector (w) multiply separate columns. Hence, the operation is not commutative.
Here is the idea with R code:
> v = c(1,7,5); w = c(2,0,3); beta=c(pi^(1/2),pi^(1/3),pi^(1/5)); gamma = c(1/3,1/5,1/7)
> sum(((beta %o% gamma) * v) %*% w) # same as sum((beta %*% t(gamma) * v) %*% w)
[1] 20.05487
> sum(((beta %o% gamma) * w) %*% v) # not a commutative operation:
[1] 17.90857
Or more simply, (\vec \beta \cdot \vec v \times \vec \gamma \cdot \vec w = 308)
[\begin{align} (\beta \otimes \gamma)[v,w]&=\langle \beta,v \rangle \times \langle \gamma,w \rangle\[2ex] & =\small \left(\color{blue}{\sqrt\pi} \times \color{magenta} 1 \quad+\quad \color{blue}{\sqrt[3]\pi} \times \color{magenta}{7} \quad +\quad \color{blue}{\sqrt[5]\pi} \times \color{magenta}5 \right) \times \left(\color{red}{\frac{1}{3}} \times \color{orange}2 \quad+\quad \color{red}{\frac{1}{5}} \times \color{orange} 0 \quad +\quad \color{red}{\frac{1}{7}} \times \color{orange} 3 \right) \[2ex] &=18.31097\times 1.095238\[2ex] &= 20.05487\end{align}]
> v = c(1,7,5); w = c(2,0,3); beta=c(pi^(1/2),pi^(1/3),pi^(1/5)); gamma = c(1/3,1/5,1/7)
> beta %*% v * gamma %*% w
[,1]
[1,] 20.05487
Does it obey bilinearity?
[(\beta\otimes \gamma)[v,w]\overset{?}=(\beta\otimes \gamma)\Bigg[\left(\frac{1}{5}v\right),\left(5,w\right)\Bigg] ]
> v_prime = 1/5 * v
> w_prime = 5 * w
> beta %*% v_prime * gamma %*% w_prime
[,1]
[1,] 20.05487 #Check!
[(\beta\otimes \gamma)[v, u + w]\overset{?}=(\beta\otimes \gamma)[v,u] + (\beta\otimes \gamma)[v,w] ]
> u = c(-2, 5, 9) # Introducing a new vector...
> beta %*% v * gamma %*% (u + w)
[,1]
[1,] 49.7012
> (beta %*% v * gamma %*% u) + (beta %*% v * gamma %*% w)
[,1]
[1,] 49.7012 #... And check!
But the evaluation of the vectors is not commutative:
v = c(1,7,5); w = c(2,0,3); beta=c(pi^(1/2),pi^(1/3),pi^(1/5)); gamma = c(1/3,1/5,1/7)
beta %*% w * gamma %*% v
## [,1]
## [1,] 17.90857
beta %*% v * gamma %*% w
## [,1]
## [1,] 20.05487
Motivating Examples:
From this Quora answer:
Any vector space imbued with an inner product has a natural (T^0_2=(\underbrace{0}_{\text{takes 0 cov’s}},\underbrace{2}_{\text{takes 2 vec’s}}))-tensor sitting there: the inner product itself: it linearly takes in a pair of vectors and spits out their inner product, an element of the base field.
Similarly, any linear transformation of a vector space acts naturally as a (T^1_1=(\underbrace{1}_{\text{takes 1 cov}},\underbrace{1}_{\text{takes 1 vec}}))-tensor.
An example of a higher-order tensor is the determinant: given any linear transformation (A), from a vector space (of dimension (n)) to itself, (\det(A)) is a (T^0_n=(\underbrace{0}_{\text{takes 0 cov’s}},\underbrace{n}_{\text{takes n vec’s}}))-tensor: (\det(A) (v_1,\cdots, v_N) = (A(v_1)) \land (A(v_2)) \land\cdots\land (A(v_N))), where “(\land)” is the fully-antisymmetrized tensor product (the “wedge” product).
And of course as others have mentioned, differential topology and geometry are littered with tensors (and tensor fields / densities).
Practically the same, but more mathy:
(\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle})Let ((\Basis_{j})_{j=1}{n}) denote the standard basis of (V = \Reals{n}) and let ((\Basis^{i})_{i=1}{n}) be the dual basis of (V{*} = (\Reals^{n})^{*}). (Where possible below, I’ve taken case to use the dummy indices (i) and (j) “globally”.)
The identity transformation (I_{n}:\Reals^{n} \to \Reals^{n}) is [ \sum_{i,j=1}{n} \delta_{i}{j}, \Basis_{j} \otimes \Basis^{i} = \sum_{j=1}{n} \Basis_{j} \otimes \Basis{j}. ] Specifically, if (v = \sum\limits_{j=1}{n} v{j} \Basis_{j}), then [ I_{n}(v) = \sum_{j=1}{n} \Basis_{j} \otimes \Basis{j}(v) = \sum_{j=1}{n} v{j}\Basis_{j} = v. ] Similarly, if (A = [a_{i}{j}]) is an (n \times n) matrix, the tensor [ T = \sum_{i,j=1}{n} a_{i}{j}, \Basis_{j} \otimes \Basis{i} \in T_{1}{1}\Reals{n} ] is the linear operator whose standard matrix is (A).
If (\Basis_{j}) is written as an (n \times 1) column matrix with a (1) in the (j)th row and (0)’s elsewhere, then (\Basis^{i}) is the (1 \times n) row matrix with a (1) in the (i)th column and (0)’s elsewhere, and the tensor product (\Basis_{j}{i} := \Basis_{j} \otimes \Basis{i}) may be denoted with ordinary matrix multiplication, the outer product of a column and a row, the (n \times n) matrix with a (1) in the ((i, j))-entry and (0)’s elsewhere.
The Euclidean inner product is [ \Brak{\ ,\ } = \sum_{i=1}{n} \Basis{i} \otimes \Basis^{i}. ] If (u) and (v) are arbitrary vectors, then [ \Brak{u, v} = \sum_{i=1}{n} \Basis{i}(u), \Basis^{i}(v) = \sum_{i=1}{n} u{i}, v^{i}. ]
If (n = 2), the determinant viewed as a bilinear function of two vectors in (\Reals^{2}) is [\begin{align*} \det &= \Basis^{1} \otimes \Basis^{2} - \Basis^{2} \otimes \Basis^{1} \in T_{2}{0} \Reals{2}; \ \det(u, v) &= \Basis^{1}(u), \Basis^{2}(v) - \Basis^{2}(v), \Basis^{1}(u) \ &= u^{1} v^{2} - u^{2} v^{1}. \end{align*}]
Similarly, if (n = 3), the ordinary cross product is
[\begin{align}&(\Basis^{2} \otimes \Basis^{3} - \Basis^{3} \otimes \Basis^{2})\otimes \Basis_{1} \ + &(\Basis^{3} \otimes \Basis^{1} - \Basis^{1} \otimes \Basis^{3})\otimes \Basis_{2} \ + &(\Basis^{1} \otimes \Basis^{2} - \Basis^{2} \otimes \Basis^{1})\otimes \Basis_{3} \in T_{2}{1} \Reals{3}. \end{align}]
From this answer on Quora:
Tensors are useful when you’ve got a whole lot of coordinates that are related to each other in some structured way. A simple vector like (\mathbf v=(1,4,2)) is an example of a tensor, but it’s simple enough so that you don’t need to understand tensors in general to understand vectors. Likewise, matrices are examples of tensors, but again, they’re best understood just as matrices. The best way to understand a mathematical construct is not individually, but in context of the set (or space) of all the constructs of that type. Rather than trying to define a number, instead define what a field of numbers is; instead of defining what a vector is, consider instead all the vectors that make up a vector space. So to understand tensors of a particular type, instead consider all those tensors of the same type together.
Covariant tensor products
The simplest tensors are vectors, so we’ll build tensors up from vector spaces, which I’ll assume we already know about. Suppose we’ve got two vector spaces (V) and (W) over a field (F). You can start with more than two, but so I don’t have to use indices to begin with, I’ll take just two. You can join them together to get their tensor product (V\otimes W) which will be another vector space. The individual elements of (V\otimes W) are named as linear combinations of elements of the form (\mathbf v\otimes\mathbf w) where (\mathbf v\in V) and (\mathbf w\in W). Since they’re linear combinations, if you have (k) scalars (a_1,\ldots,a_k) in (F), vectors (\mathbf v_1,\ldots,\mathbf v_k) in (V), and vectors vectors (\mathbf w_1,\ldots,\mathbf w_k) in (W), then (\displaystyle\sum_{i=1}^k a_i\mathbf v_i\otimes\mathbf w_i=a_1\mathbf v_1\otimes\mathbf w_1+\cdots+a_k\mathbf v_k\otimes\mathbf w_k) is a typical tensor in (V\otimes W). But there’s a requirement that we make on the symbol (\otimes), and that’s that it be linear in each coordinate. So we require that ((a_1\mathbf v_1+a_2\mathbf v_2)\otimes\mathbf w=a_1\mathbf v_1\otimes\mathbf w+a_2\mathbf v_2\otimes\mathbf w) and (\mathbf v\otimes (a_1\mathbf w_1+a_2\mathbf w_2)=a_1\mathbf v\otimes\mathbf w_1+a_2\mathbf v\otimes\mathbf w_2). That condition allows us to specify a basis for the vector space (V\otimes W) if we have bases for both (V) and (W). Suppose that (V) is a vector space of dimension (m) with basis (\mathbf b_1,\ldots,\mathbf b_m), and that (W) is a vector space of dimension (n) with basis (\mathbf c_1,\ldots,\mathbf c_n). Then (V\otimes W) is a vector space of dimension (mn) and a basis whose elements are (\mathbf b_i\otimes\mathbf c_j) where (i) varies from (1) through (m) and (j) varies from (1) through (n). That means a typical element of (V\otimes W) can be written as (\displaystyle\sum_{i=1}^m\sum_{j=1}^n a_{ij}\mathbf b_i\otimes\mathbf c_j) Rather than write this as an double sum with specified ranges for the indices, you can assume that those can be determined by context, and write the sum more simply as (\displaystyle\sum_{ij} a_{ij}\mathbf b_i\otimes\mathbf c_j) and if the bases of the component vector spaces (V) and (W) are fixed, they don’t have to be mentioned either, and the (\sum) symbol can be suppressed. That yields the greatly abbreviated notation (a_{ij}) for this tensor. Now, of course you can take a tensor product of more than two vector spaces. If you have three vector spaces with specified dimensions and bases, then a typical element of the triple tensor product would be expressed as (a_{ijk}). Although that’s a simple expression, remember that each of the three subscripts varies over a range, and it’s a triple sum where each (a_{ijk}) is a coefficient in that sum.
Dual vector spaces
If (V) is a vector space, there is a dual vector space (V^{*}) whose elements are linear transformations to the scalar field, (V\to F). If (V) has a finite basis (\mathbf b_1,\ldots,\mathbf b_n), then (V^{*}) is a vector space of the same dimension with a basis (\mathbf b_1^{*},\ldots,\mathbf b_n^{*}) where (\mathbf b_i^{*}:V\to F) is the transformation that sends the vector (a_1\mathbf b_1+\cdots+a_n\mathbf b_n) to the scalar (a_i). When writing the elements of (V) and (V^{*}) as vectors, you can write the elements of (V) as column vectors and (V^{*}) as row vectors (or vice versa).
Covariant and contravariant tensor products
These are tensor products where some of the component vector spaces are dual vector spaces. For those that are dual vector spaces, rather than using subscripts, superscripts are usually used. Take for example the tensor product (V^{*}\otimes W). A typical element would be written as (a^i_j). It is, as before, actually a sum. Here (i) is made a superscript because the tensor product is contravariant in (V), which is just another way of saying that the dual vector space (V^{*}) is being used instead of (V) itself. You can think of (a^i_j) as being a linear transformation (V) to (W), that is, a matrix. In that way, an ordinary matrix is one of these tensors where one coordinate is contravariant.
Composition of tensors
If you have two tensor products, and a vector space V appears in one of the tensor products covariantly and the other contravariantly, then you can compose them ((V^{*}\otimes W)\times(V\otimes U)\to W\otimes U) Given (a^i_j\in V^{*}\otimes W) and (b_{ik}\in V\otimes U), the result is (\mathbf\sum_ia^i_jb_{ik}\in W\otimes U). The summation sign in the result is usually suppressed so it’s written more simply as (a^i_jb_{ik}). More generally, if you have two complicated tensors where the same index appears as a superscript in one and a subscript in the other, they can be multiplied by writing them next to each other with an understood summation over that index. As an example, a matrix (V^{*}\otimes W) times a (column) vector in (V) gives a (column) vector in (W).
Preliminary definitions:
A field is a set (k) with the triad ((k, +, \cdot)) denoting two maps:
[+: k \times k \rightarrow k]
and
[\cdot: k\times k \rightarrow k]
that satisfy
CANI conditions: closure and commutativity (abelian), associativity, neutral element, inverse for every element. CANI is also satisfied by the addition and multiplication operations, except for having to remove in the multiplication case the neutral element of addition (k{0}).
A ring ((R, +, \cdot)) is similar but CANI only applies to (+). For the multiplication operation the inverses (and commutativity) gone. So every field is a ring.
Example: ((\mathbb Z, +, \cdot)) - the inverse of an inverse is not in the set.
Example: (m\times n) matrices over (\mathbb R) are not commutative, and not all of them have inverses.
A (k) ((k) is a field) vector space ((V, \color{red}{+}, \color{red}{\cdot})) has two operations defined, and different from the operations ((+) and (\cdot)) in the field, such that
[+: V \times V \rightarrow V]
and
[\cdot: k\times V \rightarrow V]
fulfilling CANI for the (\color{red}{+}) operation, but also ADDU: associativity, two distributive laws (over the (+) of the field elements; or over the (\color{red}{+}) of the vector sum operation), and scalar identity.
More formally,
- Associativity of addition
- Commutativity of addition
- Identity element of addition
- Inverse element of addition
- Compatibility of scalar multiplication with field multiplication (a(b\mathbf v)=(ab)\mathbf v)
- Identity element of scalar multiplication
- Distributivity of scalar multiplication w.r.t. vector addition
- Distributivity of scalar multiplication w.r.t. field addition
Basis of a vector space ((V, +, \cdot)):
If there is no further structure to the vector space we can only define a subset (B\in V) called Hamel basis. Its conditions are:
Every finite subset, ({b_1,\cdots,b_n}\subset B) is linearly independent. 1.
For every element (\mathbf v\in V) there exist (v^1,\cdots,v^m\in k) such that (\mathbf v=\sum_{i=1}^m v^i,b_i.)
The dimension of the vector space is the cardinality of the basis.
A module is the equivalent of a vector space over a RING, as opposed to a FIELD.
A homomorphism is a structure-preserving linear map (\large f) between two algebraic structures of the same type (such as two groups, two rings, or two vector spaces). In the case of two vector spaces, each equipped with (+) and (\cdot) operations ((V) with (\color{red}{+}) and (\color{red}{\cdot}) and (W) with (\color{orange}{+}) and (\color{orange}{\cdot})):
[f: V \rightarrow W] fulfills:
[\forall v_1, v_2 \in V: f(v_1 \color{red}{+} v_2) = f(v_1) \color{orange}{+} f(v_2)]
and
[\forall \lambda \in k, v\in V: f(\lambda \color{red}{\cdot} v) = \lambda\color{orange}{\cdot}f(v).]
A bijective linear map is called a vector space ISOMORPHISM.
Two vector spaces (V) and (W) are isomorphic (V\cong W) if (\exists) and isomorphism: (f: V\rightarrow W.)
An ENDOMORPHISM is a linear map of (V) onto itself. Hence, (\text{End}(V) := \text{Hom}(V,V).)
An AUTOMORPHISM is an invertible linear map of (V) on itself. Hence, (\text{Aut}(V) := {f: V \overset{\sim}\rightarrow V | \text{invertible}})
Automorphisms are a subset (\text{Aut}(V) \subset \text{End}(V)) of endomorphisms.
We can define the set of ALL LINEAR MAPS from (V \rightarrow W) as (\text{Hom}(V,W):= {f:V \overset{\sim}\rightarrow W}) with the (\sim) denoting linear map.
Is this set of all linear maps a vector space? Yes, by defining,
[\boxed{\color{blue}{+}}: \text{Hom}(V, W) \times \text{Hom}(V,W) \rightarrow \text{Hom}(V,W)] by taking the pair of linear functions and mapping it to
[(f,g)\mapsto f;\boxed{+};g,]
where (f;\boxed{+};g) is again a map from (V) to (W) defined by (v\mapsto f(v) \color{orange}{+} g(v))
and also defining
[\boxed{\color{blue}{\cdot}}:(\lambda,\boxed{\color{blue}{\cdot}},g) (V)= \lambda \color{orange}{\cdot} g(V)]
(V) Star, Dual Vector Space or (V^*:)
[\large V^* := \text{Hom}(V, k)]
Here (k) is considered a vector space.
So (V^*) is a set of linear functionals from a vector space to the field (k), which is considered in this context just another vector space with addition and multiplication inherited from the field operations.
Multilinear maps:
So in general, maps can be more and more complicated, and for example, [V\times V\times V^*\times V\times V^* \to \mathbb R] would be the set of all possible tensor products of the form [T_{\mu;\nu}{}{\gamma}{}_\rho{}\eta\quad e^\mu\otimes e^\nu\otimes e _\gamma\otimes e^\rho \otimes e_\eta]
with (T_{\mu;\nu}{}{\gamma}{}_\rho{}\eta) corresponding to the components of the tensor, which are the only part usually transcribed (the basis vectors are implicit). Hence, is important to keep the spaces and order of the sub- and supra-scripted Greek letters.
However, there is a system to the madness: The vectors in the tensor product come first by convention: e.g. (V\otimes V^*) as opposed to (V^* \otimes V.)
The rank of a tensor is similarly expressed as ((\text{number of vectors, number of covectors}),) so that (V\otimes V^* \otimes V^*,) symbolizes the set of all possible tensors of rank ((1,2):)
[\begin{align} T^\mu{}_{\nu\lambda}\left[e_\mu \otimes e^\nu \otimes e^\lambda\right]\left(B_\eta ,e^\eta, A^\delta,e_\delta,C^\gamma,e_\gamma\right)&=\langle B_\eta,e^\eta, e_\mu \rangle,\langle e^\nu,A^\delta e_\delta \rangle,\langle e^\lambda, C^\gamma e_\gamma\rangle\[2ex] &=T^\mu{}_{\nu\lambda};B_\eta A^\delta C^\gamma ; \langle e^\eta, e_\mu \rangle,\langle e^\nu, e_\delta \rangle,\langle e^\lambda, e_\gamma\rangle\[2ex] &=T^\mu{}_{\nu\lambda};B_\eta A^\delta C^\gamma ;\delta^\eta{}_\mu; \delta^\nu{}_\delta; \delta^\lambda{}_\gamma\[2ex] &= T^\mu{}_{\nu\lambda};B_\mu, A^\nu, C^\lambda \end{align}]
with the indexes in the last line of our choice.
A tensor (A) is a member of the (T^3_2) tensor product space, i.e. (A\in T^3_2(v),) if it is of the form (A^{\alpha\beta\gamma}{}_{\mu\nu}, \underbrace {e_\alpha\otimes e_\beta\otimes e_\gamma}_{\text{basis vecs }\in V}\otimes \underbrace{e^\mu\otimes e^\nu}_{\text{covecs }\in V^*},) meaning that it would “eat” (3) covectors and (2) vectors to produce a real number: (V^*\times V^* \times V^* \times V\times V\to \mathbb R.) So, (T^{\text{ no. vecs in }\otimes \text{ prod.}}_{\text{ no covecs in the }\otimes}) or a ((3,2))-rank tensor.
Here is an example of the “eating” vectors and covectors by a tensor:
[T^{\alpha\beta\gamma}{}_{\mu\nu}, \Big [ e_\alpha \otimes e_\beta \otimes e_\gamma\otimes e^\mu\otimes e^\nu\Big ]\left(\underbrace{B_\eta e^\eta, C_\omega e^\omega, F_\epsilon e^\epsilon}_{\text{eats 3 covectors}},\underbrace{Z^\theta e_\theta, Y^\rho e_\rho}_{\text{eats 2 vectors}}\right)\to \mathbb R]
NOTE on the difference between tensor and tensor product (from the comments by XylyXylyX):
A “tensor” is an element of a “tensor product space”, i.e. (T^i_j,(v).) A tensor product space contains elements which are the “tensor product” of vectors and covectors. So a “tensor product” is a multi linear map built using the tensor product operator ((\otimes)). So a tensor is the tensor product of some vectors and covectors. Keep in mind that a tensor product of rank ((1,0)) or ((0,1)) is not really “multi” linear, it is just “linear” but we still call it a tensor, so (V) and (V^*) are tensor product spaces but small ones. And rank ((0,0)) tensor product spaces are just real numbers.
Tensors are multilinear maps, meaning that if we multiply by a scalar any of the entries in (V\times V\times\cdots\times V^*\times V^*\times \cdots \to \mathbb R,) keeping every other component constant, the result will be multiplied by the same scalar.
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