# Reason for $(0,1)$ tensor as a vector?

Assume, we have a finite-dimensional vector space $$V = (R^3,+,-)$$ over the field $$(R,+,-)$$ and the dual space is the set of all the linear maps, $$V^* = \{f:R^3 \to R| f \text{ is linear}\}$$. We define $$(r,s)$$ tensor as multilinear map: $$V \times ... \times V(\text{r times})\times V^* \times ... \times V^*(\text{s times}) \to R$$. I understand that we can view the dual space, $$V^{*}$$ as a $$(1,0)$$ tensor since it maps $$R^3 \to R$$. But how can we view $$(0,1)$$ tensor as a vector since a $$(0,1)$$ tensor will be a map between $$V^* \to R$$.

One can construct maps $$V^*\to R$$ by evaluation of each linear functional $$f$$ in a fixed element $$v_0$$ of $$V$$, that is $$f\mapsto f(v_0),$$ such a map is linear, since for a linear combination $$Af+Bg$$ in $$V^*$$ one can see $$(Af+Bg)(v_0)=Af(v_0)+Bg(v_0),$$ where $$A,B\in R$$.

Taking different $$v_0$$'s and if you call them $$\Phi_{v_0}$$ those maps above, you can see that the set $$\{\Phi_{v_0}\}$$ is a vector space under the operations

1. $$\Phi_{v_1}+\Phi_{v_2}=\Phi_{v_1+v_2}$$ and
2. $$k\Phi_{v_0}=\Phi_{kv_0}$$ for $$k\in R$$.

A basis for this space is got by considering a basis for $$V$$: If $$\{b_1,..., b_n\}$$ is such one then $$\{\Phi_{b_1},..., \Phi_{b_n}\},$$ will give you a basis for this vector space.

• Is $\Phi$ isomorphic with the vector space $V$? Is $\Phi = \{f:V \to V^* \}$? Apr 28 at 15:37
• the isomorphism $V$ to the set $\{\Phi_{v}\}$ is $$v_0\mapsto\Phi_{v_0}$$ Apr 28 at 15:41
• Is it, $\Phi$ is the set of maps from $V \to V^{**},$$\Phi = \{f: V \to V^{**}\}$ and this $\Phi$ is isomorphic to $V$. So, we say (0,1) tensor is a vector since $\Phi$ is isomorphic to the vector space $V$. Apr 28 at 16:01
• man, I don't use the letter $\Phi$ alone, but I can tell you that $V^{**}$ is our set $\{\Phi_{v}\}$ Apr 28 at 16:09

As the vector space $$V$$ is finite-dimensional, there is an isomorphism $$\Phi:V\to V^{**}$$, given by $$[\Phi(v)](f)=f(v)$$ for $$v\in V$$ and $$f\in V^*$$.

• thanks for the answer. I am new in this, so it is hard for me to grasp at first time. It would be helpful if you can elaborate the answer. Apr 28 at 13:33

I'm a bit surprised by the terminology you're using.

In general, a tensor product of vector spaces $$A,B$$ is a vector space $$A \otimes B$$ together with a bilinear map $$\Phi \colon A \times B \to A \otimes B$$ that satisfies the universal property $$\forall \ \text{bilinear } \Psi \colon A \times B \to C \quad \exists ! \ \text{linear } \overline{\Psi} \colon A \otimes B \to C \qquad \text{such that } \Psi = \overline{\Psi} \circ \Phi.$$ One can check that such a space exists, and all possible choices are isomorphic. In this sense, we can call $$A \otimes B$$ the tensor product of $$A$$ and $$B$$.

In a similar way, one defines the tensor product of any family $$A_i$$ of vector spaces by considering multilinear maps.

Let $$V$$ be now a real vector space. The space of all $$(r,s)$$-tensors - as you describe it - is the space of all multilinear maps $$V \times ... \times V(\text{r times})\times V^* \times ... \times V^*(\text{s times}) \to \mathbb{R}.$$ According to the definition, this is isomorphic to the space of linear maps $$V \otimes ... \otimes V(\text{r times})\otimes V^* \otimes ... \otimes V^*(\text{s times}) \to \mathbb{R},$$ or in other words, to the dual space $$(V \otimes ... \otimes V(\text{r times})\otimes V^* \otimes ... \otimes V^*(\text{s times}))^*,$$ which in turn is isomorphic to $$V^* \otimes ... \otimes V^*(\text{r times})\otimes V^{**} \otimes ... \otimes V^{**}(\text{s times}).$$ In this context, the space of $$(0,1)$$-tensors is $$V^{**}$$, which in the special case of finite dimensional spaces is naturally isomorphic to $$V$$.

However, if we stick to the definition of tensor product, then we see that the tensor product of $$V$$ is just $$V$$ ($$1$$-linear maps are just linear maps). Your definition actually involves taking the dual space twice, which of course makes no difference as long as $$V^{**}=V$$.

• I think some physics texts (like Wald's GR) that want to be a little more mathy with their discussion of tensors than "something that transforms like a tensor" define them in this way as multilinear maps on the product so as to give a technically correct definition somewhat in the abstract spirit without having to write down the definition of a tensor product. Apr 28 at 22:11
• Thanks! I suspected something like this but since I only knew differential geometry books, I've never encountered this definition. Indeed, there is a certain pedagogical payoff here. Apr 29 at 8:17