# Show that $V\otimes V\simeq L(V^*,V^*,\mathbb{R})$

I am trying to understand tensor products and I would like to show that $$V\otimes V\simeq L(V^*,V^*,\mathbb{R})$$. In Lee's book about smooth manifolds is the following proof for the case $$V^*\otimes V^*\simeq L(V,V,\mathbb{R})$$:

According to the book it follows directly from $$V\simeq V^{**}$$, since everything is assumed to have finite dimension. I am fine with that. Nevertheless I was wondering if it couldn't be done as it is done in the above proof.

For $$(v,w)\in V\times V$$ and $$\omega,\eta\in V^*$$ I would define $$\Phi :V\times V\to L(V^*,V^*,\mathbb{R}),\Phi(v,w)[\omega,\eta]:=\omega(v)\cdot\eta(w)$$. I suppose it is a bilinear map. If it is, then it decents to a linear map

$$\tilde{\Phi} :V\otimes V\to L(V^*,V^*,\mathbb{R}), \tilde{\Phi}(v\otimes w)[\omega,\eta]:=\omega(v)\cdot\eta(w)$$. Now I am stuck. I am not sure what the basis vectors for $$L(V^*,V^*,\mathbb{R})$$ are or how to show that it is a bijection. I am also not sure if the apprach is okay or complete nonsense.

Thank you very much in advance!

• Conceptually, I think it is better to deal with two different (finite-dimensional) vector spaces $V$ and $W$ to see what is really happening: show $V^* \otimes W \cong {\rm Hom}(V,W)$. In the special case $W = V$, this becomes $V^* \otimes V \cong {\rm Hom}(V,V)$.
– KCd
Commented Feb 19, 2021 at 0:38

You just need to show that, $$L(V^*,V^*,\mathbb{R})$$ satisfies the universal property of tensor product.

Consider the map $$\Phi :V\times V\to L(V^*,V^*,\mathbb{R}).$$

Let, $$W$$ be a vector space with a bilinear map $$B$$ from $$V\times V$$ to $$W.$$

Suppose, $$\{v_i^*\}$$ is a basis of $$V^*$$ and if $$v^*=\sum_{i} a^i v_i^{*},w^*=\sum_{j} b^j v_j^*,$$ we have, $$B'(v^*,w^*)=\sum_{i,j} a^ib^jB'(v_i^*,v_j^*)$$

Then, there is a unique linear map $$T :L(V^*,V^*,\mathbb{R})\to W$$ defined by, $$T(B')=\sum_{i,j}a^ib^jB(v_i,v_j)$$ such that the diagram commutes.

Thus, from the universal property of tensor product, $$V\otimes V\simeq L(V^*,V^*,\mathbb{R})$$

Notation: In order to distinguish the abstract tensor product "$$\otimes$$" from the tensor product of covectors defined in example $$12.2$$, if $$Z$$ is a real vector space, and $$f_1,f_2 \in Z^*$$, let's denote by $$f_1 \,\widetilde{\otimes}\, f_2$$ to the bilinear map in $$\mathrm L(Z,Z;\mathbb R)$$ given by $$(f_1 \,\widetilde{\otimes}\, f_2)(z_1,z_2) := f_1(z_1)f_2(z_2)$$. Also, if $$z \in Z$$, let $$\operatorname{ev}_z \in Z^{**}$$ be "evaluation at $$z$$", that is, $$\operatorname{ev}_z(f) = f(z)$$ for every $$f \in Z^*$$.

Now, let $$I := \{1,\dots,\dim V\}$$, and suppose that $$\{e_i : i \in I\}$$ is a basis for $$V$$. Then proposition $$12.8$$ tells us that $$\{e_\alpha \otimes e_\beta : (\alpha,\beta) \in I \times I\}$$ is a basis for $$V \otimes V$$, and if $$\{e^i : i \in I\}$$ is the dual basis of $$\{e_i : i \in I\}$$, proposition $$12.4$$ tells us that $$\{\operatorname{ev}_{e_\alpha} \!\widetilde{\otimes} \operatorname{ev}_{e_\beta}\! : (\alpha,\beta) \in I \times I\}$$ is a basis for $$\mathrm L(V^*,V^*;\mathbb R)$$, simply because $$\{\operatorname{ev}_{e_i}\! : i \in I\}$$ is the dual basis of $$\{e^i : i \in I\}$$.

Finally, note that your map $$\widetilde{\Phi} : V \otimes V \to \mathrm L(V^*,V^*;\mathbb R)$$ sends $$e_\alpha \otimes e_\beta \in V \otimes V$$ to the bilinear map $$\widetilde{\Phi}(e_\alpha \otimes e_\beta) \in \mathrm L(V^*,V^*;\mathbb R)$$ such that $$\widetilde{\Phi}(e_\alpha \otimes e_\beta)(f_1,f_2) = f_1(e_\alpha)f_2(e_\beta) = \operatorname{ev}_{e_\alpha}(f_1) \operatorname{ev}_{e_\beta}(f_2)$$ for every $$f_1$$ and $$f_2$$ in $$V^*$$, in other words, $$\widetilde{\Phi}(e_\alpha \otimes e_\beta) = \operatorname{ev}_{e_\alpha} \!\widetilde{\otimes} \operatorname{ev}_{e_\beta}\!$$. Thus $$\widetilde{\Phi}$$ take basis elements in $$V \otimes V$$ to basis elements in $$\mathrm L(V^*,V^*;\mathbb R)$$, and this proves that $$\widetilde{\Phi}$$ is an isomorphism.

• I think that clears up a few things.Thanks!
– Hans
Commented Feb 19, 2021 at 14:08