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Note: the present entry employs the terminology and notation defined and described in the entry on tensor arrays. To keep things reasonably self contained we mention that the symbol $\tspace{p,q}$ refers to the vector space of type $(p,q)$ tensor arrays, i.e. maps $$I^p\times I^q\rightarrow \kfield,$$ where $I$ is some finite list of index labels, and where $\kfield$ is a field.
Let $p_1,p_2,q_1,q_2$ be natural numbers. Outer multiplication is a bilinear operation $$\tspace{p_1,q_1} \times \tspace{p_2,q_2} \rightarrow \tspace{p_1+p_2,q_1+q_2}$$ that combines a type $(p_1,q_1)$ tensor array $X$ and a type $(p_2,q_2)$ tensor array $Y$ to produce a type $(p_1+p_2,q_1+q_2)$ tensor array $XY$ (also written as $X\otimes Y$ , defined by $$ (XY)^{i_1\ldots i_{p_1} i_{p_1+1} \ldots i_{p_1+p_2}}_{j_1\ldots j_{q_1} j_{q_1+1} \ldots j_{q_1+q_2} } = X^{i_1\ldots i_{p_1}}_{j_1\ldots j_{q_1}} Y^{i_{p_1+1}\ldots
i_{p_1+p_2}}_{j_{q_1+1}\ldots j_{q_1+q_2}} $$ Speaking informally, what is going on above is that we multiply every value of the $X$ array by every possible value of the $Y$ array, to create a new array, $XY$ Quite obviously then, the size of $XY$ is the size of $X$ times the size of $Y$ and the index slots of the product $XY$ are just the union of the index slots of $X$ and of $Y$
Outer multiplication is a non-commutative, associative operation. The type $(0,0)$ arrays are the scalars, i.e. elements of $\kfield$ they commute with everything. Thus, we can embed $\kfield$ into the direct sum $$\bigoplus_{p,q\in\natnums} \tspace{p,q},$$ and thereby endow the latter with the structure of an $\kfield$ algebra 1.
By way of illustration we mention that the outer product of a column vector, i.e. a type $(1,0)$ array, and a row vector, i.e. a type $(0,1)$ array, gives a matrix, i.e. a type $(1,1)$ tensor array. For instance: $$ \begin{pmatrix} a \\ b \\ c \end{pmatrix}\otimes \begin{pmatrix} x & y & z \end{pmatrix} = \begin{pmatrix} ax & ay & az \\ bx & by & bz \\ cx & cy & cz \end{pmatrix} ,\quad a,b,c,x,y,z\in \kfield $$
Footnotes
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- We will not pursue this line of thought here, because the topic of algebra structure is best dealt with in the a more abstract context. The same comment applies to the use of the tensor product sign $\otimes$ in denoting outer multiplication. These topics are dealt with in the entry pertaining to abstract tensor algebra.
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