# Every element of a tensor product space can be expressed as a linear combination of tensor products

Hi. I'm studying the book written by John M. Lee on smooth manifolds, and I got a problem that is bothering me so much. Let me introduce first the definition of a tensor product space in this book:

Now let $$V_1,\ldots,V_k$$ be real vector spaces. We begin by forming the free vector space $$\mathscr{F}(V_1\times\cdots\times V_k)$$, which is the set of all finite formal linear combinations of $$k$$-tuples $$(v_1,\ldots,v_k$$) with $$v_i\in V_i$$ for $$i=1,\ldots,k$$. Let $$\mathscr{R}$$ be the subspace of $$\mathscr{F}(V_1\times\cdots\times V_k)$$ spanned by all elements of the following forms: $$(v_1,\ldots,av_i,\ldots,v_k)-a(v_1,\ldots,v_i,\ldots,v_k),$$ $$(v_1,\ldots,v_i+v_i',\ldots,v_k)-(v_1,\ldots,v_i,\ldots,v_k)-(v_1,\ldots,v_i',\ldots,v_k),$$ with $$v_j,v_j'\in V_j$$, $$i\in\{1,\ldots,k\}$$, and $$a\in\mathbb{R}$$. Define the tensor product of the spaces $$V_1,\ldots,V_k$$, denoted by $$V_1\otimes\cdots\otimes V_k$$, to be the following quotient vector space: $$V_1\otimes\cdots\otimes V_k=\mathscr{F}(V_1\times\cdots\times V_k)/\mathscr{R},$$ and let $$\Pi:\mathscr{F}(V_1\times\cdots\times V_k)\longrightarrow V_1\otimes\cdots\otimes V_k$$ be the natural projection. The equivalence class of an element $$(v_1,\ldots,v_k)$$ in $$V_1\otimes\cdots\otimes V_k$$ is denoted by $$v_1\otimes\cdots\otimes v_k=\Pi(v_1,\ldots,v_k),$$ and is called the (abstract) tensor product of $$v_1,\ldots,v_k$$. It follows from the definition that abstract tensor products satisfy \begin{align*} v_1\otimes\cdots\otimes av_i\otimes\cdots\otimes v_k= & a(v_1\otimes\cdots\otimes v_i\otimes\cdots\otimes v_k),\\ v_1\otimes\cdots\otimes(v_i+v_i')\otimes\cdots\otimes v_k= & (v_1\otimes\cdots\otimes v_i\otimes\cdots\otimes v_k)+\\ & +(v_1\otimes\cdots\otimes v_i'\otimes\cdots\otimes v_k). \end{align*}

After defining the tensor product of vector spaces, Prof. Lee said that every element of a tensor product space can be expressed as a linear combination of tensor products:

Note that the definition implies that every element of $$V_1\otimes\cdots\otimes V_k$$ can be expressed as a linear combination of elements of the form $$v_1\otimes\cdots\otimes v_k$$ for $$v_i\in V_i$$; but it is not true in general that every element of the tensor product space is of the form $$v_1\otimes\cdots\otimes v_k$$ (see Problem 12-1).

I don't know how to arrive at this conclusion. Could someone please do me a favor? Thank you.

• Does the fact that the natural projection is surjective play a role in this question? Feb 6, 2021 at 10:18
• Note that the product space you start with is set of all "finite formal linear combinations" of the products of individual vectors. So you start with linear combinations of products of vectors, and the essential from does not change, though the quotient map turns "product" to "tensor product". Feb 6, 2021 at 10:29
• Thank you, but what do you mean by "the essential from"? Feb 7, 2021 at 15:19
• Should br "form' not "from" Feb 7, 2021 at 16:00

Since $$\Pi: \mathscr{F}(V_1\times\cdots\times V_k) \to V_1\otimes\cdots\otimes V_k$$ is surjective, for every $$z \in V_1\otimes\cdots\otimes V_k$$ there is $$v \in \mathscr{F}(V_1\times\cdots\times V_k)$$ such that $$z=\Pi(v)$$. Now, as every element of $$\mathscr{F}(V_1\times\cdots\times V_k)$$ is a finite (formal) linear combination of elements in $$V_1\times\cdots\times V_k$$, there exists $$m \in \mathbb Z^+$$, $$a_1,\dots,a_m \in \mathbb R$$ and $$(v_{i1},\dots,v_{ik}) \in V_1 \times \cdots \times V_k$$ for every $$i \in \{1,\dots,m\}$$ such that $$v = \sum_{i=1}^m a_i(v_{i1},\dots,v_{ik}).$$ Then, using the linearity of $$\Pi$$ we get that $$z = \Pi(v) = \sum_{i=1}^m a_i \Pi(v_{i1},\dots,v_{ik}) = \sum_{i=1}^m a_i(v_{i1} \otimes \dots \otimes v_{ik}).$$

• May I ask a stupid question? Could you tell me the way your indices $i_1,...,i_k$ are going? I wonder if they can function well. Let's say there are like $m$ elements of $V_1\times\cdots\times V_k$ whose images under $v$ are nonzero. If I were you, I would assume those $m$ elements are of the form $(v_i^{(1)},\cdots,v_i^{(k)})$ with $i=1,\ldots,m$. Feb 8, 2021 at 5:39
• @Steve I have rewritten my answer as you suggest, I like it! Feb 8, 2021 at 6:19