# Dimension of a vector space of all tensors of rank $(k,l)$ in 4D

Dual vector space is the set of all linear functionals defined on a given vector space. The vector space and dual vector space is isomorphic and hence have the same dimension. A rank $$(k,l)$$ tensor is a multilinear functional defined on the product space consisting of $$k$$ dual spaces and $$l$$ vector spaces. Hence analogues to above,the set of all $$(k,l)$$ tensors should be isomorphic to the product space consisting of $$k$$ dual vector spaces and $$l$$ vector spaces, hence should have a dimension $$4k+4l$$, assuming we are considering the spacetime. But set of all $$(k,l)$$ rank tensors have $$4^{k+l}$$ basis vectors defined using tensor product hence have the dimension $$4^{k+l}$$. Why is that set of all tensors of rank $$(k,l)$$ is not isomorphic to the product space? How can we interpret these extra degree of freedom.

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– user237964
Commented Mar 26, 2023 at 17:59

A rank $$(k,l)$$ tensor is a multilinear functional defined on the product space consisting of $$k$$ dual spaces and $$l$$ vector spaces.

Correct (assuming you mean, as I think you really do, the Cartesian product rather than tensor product).

Hence analogues to above,the set of all $$(k,l)$$ tensors should be isomorphic to the product space consisting of $$k$$ dual vector spaces and $$l$$ vector spaces…

This is wrong! We want the dimension of the space of multilinear functionals, not the dimension of the domain of these multilinear functionals (which is what you’re calculating). We’re looking for all possible multilinear maps on a large domain space, which is why we get a potentially larger space.

I would start off with the following comment: it is usually good to reduce a multilinear algebra problem to a linear algebra one. Suppose we have three vector spaces, $$V_1,V_2,W$$, over a fixed field $$\Bbb{F}$$. Let $$\text{Hom}^2(V_1\times V_2;W)$$ be the vector space of bilinear maps $$V_1\times V_2\to W$$. There is a canonical isomorphism \begin{align} \Phi:\text{Hom}^2(V_1\times V_2;W)\to \text{Hom}(V_1,\text{Hom}(V_2,W)) \end{align} defined simply as $$\Phi(T)= [v_1\mapsto T(v_1,\cdot)]$$, i.e given a bilinear $$T$$, we define $$\Phi(T)$$ to be the linear map which assigns to each $$v_1\in V_1$$, the linear map $$T(v_1,\cdot)= [v_2\mapsto T(v_1,v_2)]$$. I leave it to you to verify that $$\Phi$$ is a linear map, and that it is invertible with inverse \begin{align} \Phi^{-1}:\text{Hom}(V_1,\text{Hom}(V_2,W))\to \text{Hom}^2(V_1\times V_2;W) \end{align} given by $$\left(\Phi^{-1}(S)\right)(v_1,v_2):=\left(S(v_1)\right)(v_2)$$.

To understand this, you just need to keep your head straight about the order of evaluation, and the idea of ‘freezing one variable’, or as is also known, currying. Therefore, the above argument shows that these two vector spaces are isomorphic, and hence have the same dimension: \begin{align} \dim \text{Hom}^2(V_1\times V_2;W)&=\dim \text{Hom}(V_1,\text{Hom}(V_2,W))\\ &=\dim V_1\cdot \dim \text{Hom}(V_2,W)\\ &=\dim V_1\cdot \dim V_2\cdot \dim W, \end{align} because the dimension of the space of linear maps between two vector spaces is the product of dimensions (the pedestrian way of thinking is that an $$m\times n$$ matrix has $$m\cdot n$$ entries which can be filled, not $$m+n$$). Again, note carefully that this is different from the dimension of the domain, $$\dim(V_1\times V_2)=\dim V_1+\dim V_2$$.

Extending this idea using induction (take $$V_1=\dots= V_k=V^*$$ and $$V_{k+1}=\cdots = V_{k+l}=V$$ and $$W=\Bbb{F}$$ the underlying field), you see that the $$(k+l)$$-fold multilinear maps form a vector space of dimension $$(\dim V)^{k+l}$$ (assuming $$\dim V$$ is finite of course, so that $$\dim V=\dim V^*$$).

It is isomorphic to the tensor product space, but the tensor product space does not have dimension $$4(k+l)$$. If you consider $$(1,1)$$-tensors, then the space $$V^* \otimes V$$ is spanned by the basis $$\big\{\epsilon^i \otimes e_j\big\}$$, where $$\{\epsilon^i\}$$ is a basis for $$V^*$$, $$\{e_j\}$$ is a basis for $$V$$, and $$i$$ and $$j$$ each take on one of $$\mathrm{dim}(V)$$ values. In 4 dimensions, that means 16 basis elements for $$V^*\otimes V$$, not 8.

• In 'linear algebra done right' by Sheldon Axler, there is a theorem which states that dimension of product of vector spaces is the sum of dimensions Commented Mar 14, 2023 at 4:11
• @Grace Are you sure that the theorem doesn't refer to the dimension of a direct sum of vector spaces? $\mathrm{dim}(V\otimes W) = \mathrm{dim}(V) \cdot \mathrm{dim}(W)$, but $\mathrm{dim}(V\oplus W) = \mathrm{dim}(V) + \mathrm{dim}(W)$. Commented Mar 14, 2023 at 4:26
• @J.Murray seems to be a language barrier here: it is also common to refer to the (external) direct sum as the (direct/Cartesian) product of vector spaces, and as we know, in the case of finitely many copies of a spaces, the external direct sum is equal to the direct product. So, for clarity, I’d suggest adding the adjective ‘tensor’ infront of product space in your answer. Commented Mar 14, 2023 at 6:45
• @peek-a-boo Good point, edited. Commented Mar 14, 2023 at 6:52