>>11832322

>>11832398

Now given some vectors [math] v_1,...,v_r\in V [/math] and [math] v_1^*,...,v_s^*\in V^*, [/math] we can define an [math] (r,s) [/math] tensor as [math] v_1\otimes...\otimes v_r\otimes v_1^*\otimes ...\otimes v_s^*, [/math] which acts on an [math] (r,s) [/math] tuple (in post above) as [eqn] v_1\otimes...\otimes v_r\otimes v_1^*\otimes ...\otimes v_s^*(w_1^*,...,w_r^*,w_1,...,w_s)=w_1^*(v_1)...w_r^*(v_r)v_1^*(w_1)...v^*(s)(w_s), [/eqn] recalling that functionals act on vectors to give a scalar. Now an [math] (r,s) [/math] tensor needn't act only on an [math] (r,s) [/math] tuple. It can act on any tensor - for example [eqn] v_1\otimes...\otimes v_r\otimes v_1^*\otimes ...\otimes v_s^*(w_1,...,w_s)=v_1^*(w_1)...v_s^*(w_s)v_1\otimes ...\otimes v_r. [/eqn] There we see an [math] (r,s) [/math] tensor acting on a [math] (0,s) [/math] tuple (or a [math] (s,0) [/math] tensor - see below) to give us a [math] (r,0) [/math] tensor. Hopefully you can see what rank you get when it acts on an arbitrary tensor.

Now just note that the set of tuples [math] (w_1^*,...,w_r^*,w_1,...,w_s) [/math] has a one to one correspondence with the set of [math] (s,r) [/math] tensors, [math] w_1\otimes ...\otimes w_s\otimes w_1^*\otimes ...\otimes w_r^*, [/math] which shows why it is natural to allow tensors to act on arbitrary tensors.