Definition of tensor as a map to $\mathbb{R}$ and type (1,1)

Question

I am a little confused about a definition of tensors. If I define a space of all tensors as: $$T_l^k(V) \equiv \underbrace{V\otimes\ldots\otimes V}_l\otimes \underbrace{V^*\otimes \ldots \otimes V^*}_k$$, a tensor is then a member of this space and they are defined as a multilinear maps from $\underbrace{V^* \times \ldots V^*}_l \times \underbrace{V \times \ldots\times V}_k$ (or I think more precisly, it should be written $V^* \times \ldots V^* \times {V^*}^* \times \ldots\times {V^*}^*$) to $\mathbb{R}$.

Thus for each set of $(v_1^*,\ldots,v_l^*,v_1,\ldots,v_k)\in V^* \times \ldots V^* \times V \times \ldots\times V$, $T(v_1^*,\ldots,v_l^*,v_1,\ldots,v_k)$ is a number.

But then my notes give examples of various obvious tensors like vectors, linear forms, bilinear forms. But then it says (also wikipeadia and just any other source i can find) that the tensor of type $(1,1)$ is a linear map from $V$ into self: $f:V\to V$ but something is wrong here because the tensor should produce a number not map a vector space onto itself. Or is this the same situation as with double dual space that people write $V$ instead of ${V^*}^*$ even though that is like comparing apples and pork, they are just completely different thing (even though there is an isomorphism between them)

Answer 1

The key point is the duality. I assume that your vector space $V$ is finite dimensional : therefore, an element $v \in V$ is nothing but an element of $(V^*)^* = \text{Hom}(V^*, \mathbb R)$. So any element in $T^k_l(V)$ can be though as a linear map. Tensoring is just multypling map together, so we get a linear map $f : \underbrace{V^* \otimes\ldots\otimes V^*}_l\otimes \underbrace{V\otimes \ldots \otimes V}_k \to \mathbb R$.

For example, an element $v \otimes f$ where $v \in V, \gamma \in V^*$ is exactly a map $g : V \to V, w \mapsto \gamma(w)v$.

As another example, a bivector $v \otimes w \in V \otimes V$ is exactly a map $h : {V^* \otimes V^* \to \mathbb R, \alpha \otimes \beta \mapsto \alpha(v)\beta(w)}$.

Answer 2

When we say that a linear map $f\colon V \to V$ is a $(1,1)$ tensor, we are identifying $f$ with the $(1,1)$ tensor $\widetilde{f}\colon V^\ast\times V \to \Bbb R$ given by $$\widetilde{f}(\phi,v) = \phi(f(v)).$$The point is: fix a basis $(e_1,\ldots,e_n)$ for $V$. Then $$f(e_j)=\sum_{i=1}^n a_j^ie_i.$$If $(e^1,\ldots, e^n)$ is the dual basis, then $$\widetilde{f}(e^i,e_j) = e^i(f(e_j)) = e^i\left(\sum_{k=1}^na_j^k e_k\right) = \sum_{k=1}^na_j^ke^i(e_k) = a_j^i,$$whence $$\widetilde{f} = \sum_{i,j=1}^n a_j^ie^j\otimes e_i.$$So they're essentially the same thing: both are represented by the numbers $a_j^i$.