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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 $V^* \to R$.

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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.

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  • $\begingroup$ Is $\Phi$ isomorphic with the vector space $V$? Is $\Phi = \{f:V \to V^* \}$? $\endgroup$ Apr 28 at 15:37
  • $\begingroup$ the isomorphism $V$ to the set $\{\Phi_{v}\}$ is $$v_0\mapsto\Phi_{v_0}$$ $\endgroup$
    – janmarqz
    Apr 28 at 15:41
  • $\begingroup$ Is it, $\Phi$ is the set of maps from $V \to V^{**}, $$\Phi = \{f: V \to V^{**}\}$ and this $\Phi$ is isomorphic to $V$. So, we say (0,1) tensor is a vector since $\Phi$ is isomorphic to the vector space $V$. $\endgroup$ Apr 28 at 16:01
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    $\begingroup$ man, I don't use the letter $\Phi$ alone, but I can tell you that $V^{**}$ is our set $\{\Phi_{v}\}$ $\endgroup$
    – janmarqz
    Apr 28 at 16:09
  • $\begingroup$ glad to be helpful... merci beaucoup! $\endgroup$
    – janmarqz
    Apr 29 at 14:03
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As the vector space $V$ is finite-dimensional, there is an isomorphism $\Phi:V\to V^{**}$, given by $$[\Phi(v)](f)=f(v)$$ for $v\in V$ and $f\in V^*$.

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  • $\begingroup$ 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. $\endgroup$ Apr 28 at 13:33
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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 $V^{**}=V$.

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    $\begingroup$ 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. $\endgroup$
    – jawheele
    Apr 28 at 22:11
  • $\begingroup$ 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. $\endgroup$ Apr 29 at 8:17

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