Is a pure tensor simply a quotient of vector tuples?

I am familiarizing myself with the notion of a tensor product. I believe I have a clear idea on what the tensor product itself is. That is, for vector spaces $$V_1, \ldots, V_k$$ $$V_1 \otimes \cdots \otimes V_k = \Big\{\sum_1^n a_i (v_1, \ldots, v_k) : v_i \in V_i, a_i \in \mathbb{F}, n \in \mathbb{N}\Big\}\Big/Q$$ where $$\mathbb{F}$$ is some scalar field and $$Q$$ is a set we quotient out to obtain the desired properties of a product (scalar multiplication, linearity, distributivity).

If I recall correctly, I have also seen tensor products defined as $$V_1 \otimes \cdots \otimes V_k = \Big\{\sum_1^n a_i(v_1 \otimes \cdots \otimes v_k) : v_i \in V_i, a_i \in \mathbb{F}, n \in \mathbb{N}\Big\}$$ where the operation $$v_1 \otimes \cdots \otimes v_k$$ is defined to satisfy the desired multiplication properties we previously quotiented out.

What I am confused about is what exactly is the operation $$v_1 \otimes \cdots \otimes v_k$$ (I believe these are called pure tensors), defined on individual vectors as opposed to spaces? It seems one is used to define the other, and so the circular definition is causing me some trouble. Is $$v_1 \otimes \cdots \otimes v_k = (v_1, \ldots, v_k)\big/Q$$ where $$v_i \in V_i$$? However it wouldn't make sense to quotient out a single element, so what exactly do we mean by the image of $$(v_1, \ldots, v_k)$$ (which is said to equal $$v_1 \otimes \cdots \otimes v_k)$$ under this quotient?

Another definition is from Lee's text on smooth manifolds, where he states $$V_1 \otimes \cdots \otimes V_k = F(V_1 \times \cdots \times V_k)\Big/Q$$ where $$F$$ is the free vector space on $$V_1 \times \cdots \times V_k$$ (defined as the set of all formal linear combinations of elements of $$V_1 \times \cdots \times V_k$$). He defines $$\Pi: F(V_1 \times \cdots \times V_k) \rightarrow V_1 \otimes \cdots \otimes V_k$$ to be the so called "natural projection". Using this idea he says the equivalence class of an element $$(v_1, \ldots, v_k)$$ in $$V_1 \otimes \cdots \otimes V_k$$ is denoted by $$v_1 \otimes \cdots \otimes v_k = \Pi(v_1, \ldots, v_k).$$ What exactly is the natural projection here, I have not seen this terminology before.

As an example, let us take $$v_1 = (1,0) \in \mathbb{R}^2$$ and $$v_2 = (1, 0, 0) \in \mathbb{R}^3$$. How would one explicitly calculate this tensor product?

• Tensor products of vector spaces (over fields) are actually relatively straight-forward, if you're willing to go the the coordinate-ful (as opposed to coordinate-free) route by choosing bases for the vectors spaces for which you are taking the tensor product. This viewpoint ignores and skirts around the definition in terms of the quotient by generating a direct isomorphism between the tensor product and a vector space whose dimension is the product of the dimensions of the component vector spaces... Jun 23, 2022 at 1:42
• ...Unfortunately, this doesn't nicely generalize to tensor products of arbitrary algebraic structures (modules, rings, abelian groups, etc.) Jun 23, 2022 at 1:42
• For instance you can interpret your product $v_1\otimes v_2$ at the end of the post as $(1,0,0,0,0,0) \in \mathbb{R}^6$. Generalizing, forming the six products of the vectors in the standard bases of $\mathbb{R}^2$ and $\mathbb{R}^3$ generates the standard basis on $\mathbb{R}^6$. And so $\mathbb{R}^2\otimes\mathbb{R}^3$ is isomorphic to $\mathbb{R}^6$. Jun 23, 2022 at 1:45
• @march How do you deduce that $v_1 \otimes v_2 = (1, 0, 0, 0, 0, 0) \in \mathbb{R}^6$? Jun 23, 2022 at 2:13
• I found this write-up by Timothy Gowers to be pretty useful. Jun 23, 2022 at 16:34

2 Answers

$$\otimes$$, also called tensing, is something you get bundled with the tensor product that you don't have in an ordinary vector space. How the tensor product vector space and tensing work together are what the real "meat" behind the tensor product is. Constructions are not "the real meaning", because there are an infinite number of them that will do the job - they're really better understood as first, proofs that the tensor product exists, and second, encodings of the tensor product in the medium of sets, similar to how that, on a computer, ASCII is an encoding of text in binary numbers. The same applies to constructions of most other mathematical objects using sets.

Hence, what $$v \otimes w$$ "is" will depend on which construction you choose. In the first case, it is not circular: we define $$v \otimes w$$ to be the cell in $$Q$$ containing the ordered pair $$(v, w)$$. And in general cases, that is the

The "real meaning" behind the tensor product, and that nifty little tensing operation in comes with, is that it provides a space which lets you work with bilinear maps (generically, $$n$$-linear maps) as though they were unilinear maps. Now, I suppose you (or some others) might be thinking, "but isn't $$V \times W$$ a vector space? So isn't a bilinear map $$f: V \times W \rightarrow Z$$, a linear map from an ordered pair $$(v, w)$$, viewed as a single vector in $$V \times W$$?" Yes, it is, but remember that a bilinear map must be linear in each argument individually, and this gives them more structure that is not captured by a simple linear map out of $$V \times W$$.

Hence the tensor product. We can think of this as enriching the domain so that, in this new domain, which we call $$V \otimes W$$, being unilinear now carries all the structural weight of being bilinear on the $$V \times W$$ domain.

In particular, the tensor product as the property that every bilinear map $$f: V \times W \rightarrow Z$$, can be understood uniquely as a unilinear map $$f_\otimes : V \otimes W \rightarrow Z$$, where

$$f_\otimes(v \otimes w) := f(v, w).$$

Moreover, every vector space that has this property is isomorphic to the tensor product. The construction, then, simply shows that this is not a vacuous statement, i.e. that we are actually talking about a real mathematical object here. In this regard, it's kind of like the various constructions of the real numbers: the real numbers are "really" the single object known as "the Dedekind-complete ordered field" - what those constructions do is they prove that such a thing actually exists.

In this setting, the meaning of $$v \otimes w$$ is that it's a "package" that wraps together $$v$$ and $$w$$ into a single matrovector for processing into a linear map in such a fashion that said linear maps acquire all the extra structure bilinear maps have, which simply taking an ordered pair would not be able to do.

For finite dimensional vector spaces, you can view tensors as multilinear maps. The tensor product $$v_1 \otimes v_2$$ is the bilinear map on $$V_1^* \times V_2^*$$ with $$(v_1 \otimes v_2)(\omega_1, \omega_2) = v_1(\omega_1)v_2(\omega_2) = \omega_1(v_1)\omega_2(v_2).$$ In your specific case, $$v_1 \otimes v_2$$ is the already in simplest form as $$v_1$$ and $$v_2$$ are basis vectors.