# Understanding the notation in the definition of a tensor through an example.

A $(p,q)$ tensor, $T$ is a MULTILINEAR MAP that takes $p$ copies of $V^*$ and $q$ copies of $V$ and maps multilinearly (linear in each entry) to $k:$

$$T: \underset{p}{\underbrace{V^*\times \cdots \times V^*}}\times \underset{q}{\underbrace{V\times\times \cdots V\times V}} \overset{\sim}\rightarrow K\tag 1$$

The $(p,q)$ TENSOR SPACE is defined as a set:

\begin{align}T^p_q\,V &= \underset{p}{\underbrace{V\otimes\cdots\otimes V}} \otimes \underset{q}{\underbrace{V^*\otimes\cdots\otimes V^*}}:=\{T\, |\, T\, \text{ is a (p,q) tensor}\}\tag2\\[3ex]&=\{T: \underset{p}{\underbrace{V^*\times \cdots \times V^*}}\times \underset{q}{\underbrace{V\times \cdots \times V}} \overset{\sim}\rightarrow K\}\end{align}\tag3

is the set of all tensors where $T$ is (p,q), equipped this with pointwise addition and s-multiplication.

I can't find an example online to get an idea of what these expressions mean. I have followed, for example, all 25 lectures on tensors on this series, but these expressions are not even mentioned. I'd like to see an example that is not completely trivial, and that it could be have been dealt with using linear algebra - something with "arrow vectors" and matrices, perhaps, so that the linear functional(s) in $V^*$ and the vectors in $V$ are clearly spelled out, together with the operations entailed ($\otimes$).

If asking for an example is not a good question, a step-by-step translation in English of what these expressions are saying would be great.


• 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*}

• Thank you. I'll have to read it several times over to get to the core of it... Feb 11 '17 at 13:30
• 1. The tensor product of a vector and a covector may be viewed as a vector-valued linear function of one vector variable. In the second bullet point, $$T(v) = \sum_{i,j=1}^{n} a_{i}^{j} \Basis_{j} \otimes \Basis^{i}(v) = \sum_{i,j=1}^{n} a_{i}^{j} v^{i} \Basis_{j}.$$2. The sum you mention is the $n \times n$ matrix whose entries are all $1$. Feb 12 '17 at 0:17
• Thanks, still far from "getting it", but the last comment helps... I guess for $\mathbb R^2$ the standard basis is $\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix}\}$, and the covectors are $\{\begin{bmatrix}1&0\end{bmatrix}, \begin{bmatrix}0&1\end{bmatrix}\}$ so that... Feb 12 '17 at 0:34
• $\sum_{i,j=1}^2 \mathbf e_j\otimes e^i = \begin{bmatrix}1\\0\end{bmatrix} \otimes \begin{bmatrix}1 & 0\end{bmatrix} + \begin{bmatrix}1\\0\end{bmatrix} \otimes \begin{bmatrix}0&1\end{bmatrix} + \begin{bmatrix}0\\1\end{bmatrix} \otimes\begin{bmatrix}1&0\end{bmatrix}+\begin{bmatrix}0\\1\end{bmatrix} \otimes \begin{bmatrix}0&1\end{bmatrix}= \begin{bmatrix}1&0\\0&0\end{bmatrix}+ \begin{bmatrix}0&1\\0&0\end{bmatrix}+\begin{bmatrix}0&0\\1&0\end{bmatrix}+\begin{bmatrix}0&0\\0&1\end{bmatrix}$... Feb 12 '17 at 0:34
• Both your preceding comments are exactly correct. Consequently, including coefficients $a_{i}^{j}$ in the sum has the effect of putting $a_{i}^{j}$ into the $(i, j)$-entry of a matrix. Feb 12 '17 at 11:48