Direct Sum vs. Direct Product vs. Tensor Product [closed]

There are a lot of questions like this all over the site, but I cannot find one that resolved my confusion- what are the formal definitions of direct sums, direct products, and tensor products (in the most general sense), and how are they different?

Answer

I won’t even attempt to be the most general with this answer, because I admit, I do not have a damn clue about what perverted algebraic sets admit tensor products, for example, so I will stick with vector spaces, but I am quite sure everything I say about vector spaces works for finitely generated modules over commutative rings as well. And also that the basic concept for all direct sums and tensor products are the same, just the algebraic structures involved are different.

Direct product

I’m pretty sure the direct product is the same as Cartesian product. If $X$ and $Y$ are two sets, then $X\times Y$, the Cartesian product of $X$ and $Y$ is a set made up of all ordered pairs of elements of $X$ and $Y$.

Ergo, if $x\in X$ and $y\in Y$, then $(x,y)\in X\times Y$.

Direct sum

If $V$ and $W$ are vector spaces over a field $\mathbb{F}$, then their direct sum $V\oplus W$ is the vector space whose elements are ordered pairs of elements of $V$ and $W$, equipped with the following linear structure: $$ (x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2) \\ \alpha(x,y)=(\alpha x,\alpha y), $$ where $x,x_1,x_2\in V,\ y,y_1,y_2\in W$ and $\alpha\in\mathbb{F}$.

How is this different from the direct product? Well, the direct product can be made between arbitrary sets, and has nothing to do with algebraic properties, while the direct sum also carries over the linear structure.

However, most of the time, when you take direct product of vector spaces, you assume quietly, or directly state the existence of this linear structure on the product space, and there is one other difference:

If you take a Cartesian product of infinite amount of vector spaces, then an element of this product space can have infinite nonzero “components”, while if you take the direct sum of infinite amount of vector spaces, then an element of that sum will always have finite nonzero “terms”.

Tensor product

Oh man, this is quite different.

Basically, if you check the properties of a direct sum, you’ll see that it indeed kinda does behave like a sum. Also for example, the map that maps $x\in V$ and $y\in W$ to $(x,y)\in V\oplus W$, is absolutely not a bilinear map, as evidenced by the fact that $\alpha(x,y)\neq(\alpha x,y)$.

Question: Would it be possible to create a third vector space from $V$ and $W$ in a way, that the “pair forming map” is bilinear, and thus this new space behaves kinda like a product?

Another, seemingly unrelated question: If we have a multilinear map from a direct sum of vector spaces, then is it possible to somehow represent this map as a linear map, and have this representation be unique?

Although the two questions are seemingly unrelated, if the answer to both questions is yes, then maybe our multilinear map would appear as the composition of the “pair forming map” with a unique linear map?

These musings lead us to the actual concept of tensor product.

Definition:

Let $V$ and $W$ be finite dimensional vector spaces over the field $\mathbb{F}$. In addition, let $U$ and $X$ also be finite dimensional vector spaces over $\mathbb{F}$.

Let $p:V\times W\rightarrow X$ be a multilinear map. The pair $(X,p)$ is the tensor product of $V$ and $W$ if for every multilinear map $A:V\times W\rightarrow U$, there exists a unique linear map $A^\otimes:X\rightarrow U$ such that $$ A=A^\otimes\circ p. $$ ——————–

We usually denote $p(x,y)$ as $x\otimes y$, and $X$ as $V\otimes W$.

This also works for any amount of vector spaces, but I chose to roll with two (same goes for direct sum and direct product too).

After this, it is customary to prove existence and unicity for the tensor product. Unicity in this case means that if there is another tensor product of $V$ and $W$, then the two tensor product spaces are canonically isomorphic, and the $p$ map is given by the composition of the other “$p$ map” and the isomorphism.

Existence is usually given by constructing a tensor product directly. One elegant example is to assume that $U$ is $\mathbb{F}$ (we can do that without loss of generality, since the space of multilinear maps $V\times W\rightarrow U$ is canonically isomorphic to the space of multilinear functionals $V\times W\times U^*\rightarrow\mathbb{F}$.), and take the tensor product space to be the dual of the space of multilinear functionals over $V\times W$.

A more immediately understandable construction would be to take the tensor product space to be the space of multilinear functionals over $V^*\times W^*$, since the two are canonically isomorhpic.

Also as a closing remark, note that unlike the direct sum, which always consists of pairs $(x,y)$, elements of the tensor product space are generated by pairs $x\otimes y$, but they are not all pairs.

Attribution
Source : Link , Question Author : Community , Answer Author : Bence Racskó

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