Notes:Concrete tensor product

Notes

Let [ilmath](V_1,\mathbb{F})[/ilmath] and [ilmath](V_2,\mathbb{F})[/ilmath] both be vector spaces over the field [ilmath]\mathbb{F} [/ilmath]. Let [ilmath]V_i^*:\eq L(V_i,\mathbb{F})[/ilmath] - the dual vector space. We wish to show that:

• [ilmath]V_1\otimes V_2\cong L(V_1,V_2;\mathbb{F})[/ilmath], with some intermediate steps.

Concrete tensor "operation"

This is where I am struggling - I need to make this formal.

• Define [ilmath]\otimes_{V_1,V_2}:V^*_1\times V^*_2\rightarrow L(V_1,V_2;\mathbb{F})[/ilmath] by [math]\otimes_{V_1,V_2}:(\alpha,\beta)\mapsto\left(\begin{array}{cc}\big(:V_1\times V_2\rightarrow\mathbb{F}\big)\\\big(:(u,v)\mapsto \alpha(u)\beta(v)\big)\end{array}\right)[/math] where [ilmath]\alpha(u)\cdot\beta(v)[/ilmath] is just the multiplication operation of the field [ilmath]\mathbb{F} [/ilmath]

Is this a surjection?

• Let [ilmath]f\in L(V_1,V_2;\mathbb{F})[/ilmath] be given.
  • We claim: [ilmath]\exists (\alpha,\beta)\in V^*_1\times V^*_2[\alpha\otimes_{V_1,V_2}\beta\eq f][/ilmath] - really I cannot see how to pick [ilmath](\alpha,\beta)[/ilmath] and go from there. So we must construct it....
    • If [ilmath]f\eq g[/ilmath] for some functions of the form [ilmath](:X\rightarrow Y)[/ilmath] then [ilmath]\forall x\in X[f(x)\eq g(x)][/ilmath] - we must do this.
    • Let [ilmath](u,v)\in V_1\times V_2[/ilmath] be given - we need to show [ilmath](\alpha\otimes\beta)(u,v)\eq f(u,v)[/ilmath] (using [ilmath]\otimes[/ilmath] as short for [ilmath]\otimes_{V_1,V_2} [/ilmath])
      • Note that [math]u\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}e^{(1)}_{c_1} [/math] and [math]v\eq\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2} [/math] for [ilmath]n_i:\eq\text{Dim}(V_i)[/ilmath]
      • Then [math]f(u,v)\eq f\left(\sum^{n_1}_{c_1\eq 1}u_{c_1}e^{(1)}_{c_1},\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)[/math]

        [math]\eq\sum^{n_1}_{c_1\eq 1}u_{c_1} f\left(e^{(1)}_{c_1},\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)[/math]

        [math]\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}\sum^{n_2}_{c_2\eq 1} v_{c_2} f(e^{(1)}_{c_1},e^{(2)}_{c_2})[/math]

        [math]\eq\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}f(e^{(1)}_{c_1},e^{(2)}_{c_2})[/math]

      • and [math](\alpha\otimes\beta)(u,v)\eq\alpha(u)\beta(v)[/math]

        [math]\eq\alpha\left(\sum^{n_1}_{c_1\eq 1}u_{c_1}e^{(1)}_{c_1}\right)\beta\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)[/math]

        [math]\eq\left(\sum^{n_1}_{c_1\eq 1}u_{c_1}\alpha(e^{(1)}_{c_1})\right)\beta\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)[/math]

        [math]\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}\alpha(e^{(1)}_{c_1})\beta\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}e^{(2)}_{c_2}\right)[/math]

        [math]\eq\sum^{n_1}_{c_1\eq 1}u_{c_1}\alpha(e^{(1)}_{c_1})\left(\sum^{n_2}_{c_2\eq 1}v_{c_2}\beta(e^{(2)}_{c_2})\right)[/math]

        [math]\eq\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}\alpha(e^{(1)}_{c_1})\beta(e^{(2)}_{c_2})[/math]

      • Observe: [math]\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}f(e^{(1)}_{c_1},e^{(2)}_{c_2})\eq\sum^{n_1}_{c_1\eq 1}\sum^{n_2}_{c_2\eq 1}u_{c_1}v_{c_2}\alpha(e^{(1)}_{c_1})\beta(e^{(2)}_{c_2})[/math] is what we require