# The Unapologetic Mathematician

## More on tensor products and direct sums

We’ve defined the tensor product and the direct sum of two abelian groups. It turns out they interact very nicely.

One thing we need is another fact about the tensor product of abelian groups. If we take three abelian groups $A$, $B$, and $C$, we can form the tensor product $A\otimes B$, and then use that to make $(A\otimes B)\otimes C$. On the other hand, we could have started with $B\otimes C$ and then built $A\otimes(B\otimes C)$. If we look at the construction we used to show that tensor products actually exist we see that these two groups are not the same. However, they are isomorphic.

To see this, let’s make a bilinear function from $(A\otimes B)\times C$ to $A\otimes(B\otimes C)$. By our construction, any element of $A\otimes B$ can be represented as a sum $\sum\limits_i a_i\otimes b_i$, so linearity says we just need to consider elements of the form $a\otimes b$. Define $f(a\otimes b,c)=a\otimes(b\otimes c)$. This induces a unique linear function given by $\bar{f}((a\otimes b)\otimes c)=a\otimes(b\otimes c)$ and extending to sums of such elements. Similarly we get a linear function $\bar{f}^{-1}(a\otimes(b\otimes c))=(a\otimes b)\otimes c)$, so we have an isomorphism of abelian groups. We can thus (somewhat) unambiguously talk about “the” tensor product $A\otimes B\otimes C$.

Now let’s take a collection of abelian groups $A_i$ with $i$ running over an index set $\mathcal{I}$, and let $B$ be any other abelian group. We want to consider the tensor product
$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$

Since the direct sum is a direct product of groups, it comes with projections $\pi_k:\bigoplus_i A_i\rightarrow A_k$. Since the free product is in general a subgroup of the direct sum (a proper subgroup when the index set is infinite), we also have injections $\iota_k:A_k\rightarrow\bigoplus_i A_i$ coming from the free product. We can use these to build homomorphisms
$\iota_i\otimes1_B:A_i\otimes B\rightarrow\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$
applying $\iota_i$ to $A_i$ and the identity to $B$. By the universal property of direct sums (the one it gets from free products of groups) this gives us a homomorphism
$\alpha:\bigoplus\limits_{i\in\mathcal{I}}(A_i\otimes B)\rightarrow\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B$

On the other hand, for each $k$ we have a bilinear function sending $(a,b)$ in $\left(\bigoplus_i A_i\right)\times B$ to $\pi_k(a)\otimes b$ in $A_k\otimes B$. By the universal properties of tensor products this gives a linear function $\left(\bigoplus_i A_i\right)\otimes B\rightarrow A_k\otimes B$. The universal property of direct sums (the one it gets from direct products of groups) gives us a linear function
$\beta:\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes B\rightarrow\bigoplus\limits_{i\in\mathcal{I}}(A_i\otimes B)$

Now there’s a lot of juggling of functions and injections and projections here that I really don’t think is very illuminating. The upshot is that $\alpha$ and $\beta$ are inverses of each other, giving us an isomorphism of the two abelian groups. There’s nothing really special about the left side of the tensor product either. A similar result holds if the direct sum is the right tensorand. We can even put them together to get the really nice isomorphism:

$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes\left(\bigoplus\limits_{j\in\mathcal{J}}B_j\right)\cong\bigoplus\limits_{(i,j)\in\mathcal{I}\times\mathcal{J}}(A_i\otimes B_j)$

Neat!

April 17, 2007 - Posted by | Abelian Groups, Algebra, Group theory

1. “More on tensor products and direct sums”:

The direct sum is a co-product, not a direct product. It therefore comes with injections, not with projections. It seems to me that the proof given here might be incorrect.

Shlomi

Comment by Shlomi | July 31, 2010 | Reply

2. The direct sum is both a product and a coproduct, and it comes with both injections and projections. As I say in the second link in the post, “When we restrict our attention to abelian groups, direct products and free products are the same thing”.

See also here, keeping in mind that an Abelian group is a $\mathbb{Z}$-module.

Comment by John Armstrong | July 31, 2010 | Reply

3. I also find this proof troublesome. I believe you are confusing the terms “direct sum” and “direct product”, and unnecessarily introducing the term “free product” (I assume you mean “free product of abelian groups”, in which case this would by definition be identical to the direct sum). As you are aware, the direct sum is a subgroup of the direct product, and only equal when $\mathcal{I}$ is finite. Thus, your construction of the map $\beta$ by arguing that a collection of maps to each $A_i\otimes B$ induces a map to $\bigoplus_{i\in\mathcal{I}}(A_i\otimes B)$ is incorrect when $\mathcal{I}$ is infinite.

I’m not disputing the theorem itself – the tensor product *does* distribute over arbitrary direct sums – but I believe your proof is flawed. For an example of a correct proof, see Theorem 5.4 on p.22 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf

I also wanted to say I have used your site many times to understand things better, and you have an impressive command of an extremely broad collection of areas in math. I don’t intend to sound like I’m claiming you don’t know what you’re doing; I just think you made an honest mistake here.

Many thanks for your continued blathing,

Zev

Comment by Zev Chonoles | February 13, 2011 | Reply

• The direct sum of Abelian groups plays two roles that come from group theory, each of which gives different properties. I refer to those names (which were defined earlier and are available through tracing backlinks) to emphasize and use those properties.

Comment by John Armstrong | February 13, 2011 | Reply

• But the direct sum only plays those two roles, both product and coproduct, in the case of finitely many factors. Shlomi’s comment is not quite right; a direct sum $A=\bigoplus A_i$ (even one with infinitely many factors) does, of course, come with projections $\pi_i$ to each factor $A_i$. However, these projections do not serve to make $A$ the product of the $A_i$ if there are infinitely many $A_i$. In other words, given an infinite collection of (non-trivial) abelian groups $A_i$, their direct sum $A$ *does not* have the property that any collection of maps of abelian groups $f_i:X\rightarrow A_i$ factors uniquely through a map $f:X\rightarrow A$.

For example, let $I$ be any infinite set, let each $A_i=\mathbb{Z}$, let $X=\mathbb{Z}$, and let all the maps $f_i:X\rightarrow A_i$ be the identity map of $\mathbb{Z}$. But there is no map $f:X\rightarrow A$ such that $f_i=\pi_i \circ f$, because where could $f$ send $1\in X$ to? Not $(1,1,\ldots)$, because that is not an element of the direct sum.

Comment by Zev Chonoles | February 13, 2011 | Reply

• I would point to the Wikipedia page on biproducts, http://en.wikipedia.org/wiki/Biproduct, but it is, unhelpfully, poorly worded about this very issue. It is not made explicitly clear that when they say “In the category of abelian groups, biproducts always exist and are given by the direct sum.”, they are only referring to biproducts of finitely many factors. However, this is alluded to in other parts of the page.

Comment by Zev Chonoles | February 13, 2011 | Reply

• Let me clarify the first sentence of my antepenultimate post. I should say, “The direct sum always plays the role of the coproduct; it only additionally plays the role of the product in the case of finitely many factors.”

Comment by Zev Chonoles | February 13, 2011 | Reply

4. Zev has decided to passive-aggressively demand that I pay attention to him by complaining to me on Formspring.

So yes, Zev, go through and mentally add a finiteness condition. If you want to fill in the gap in the infinite case, start a weblog of your own and write up the proof.

I’m sorry that I don’t instantly bend to your whims, and that I have other projects and a day job that keep me from doing so. If you want more responsiveness, pay for my time.

Comment by John Armstrong | February 27, 2011 | Reply

• I truly do apologize if I seemed passive-aggressive; I don’t remember my phrasing, but I’m sure it could have been much better if I made you angry. However, your response to my (quite kindly worded) initial post indicated that you thought I was in error. Indeed, you were rather dismissive of both me and Shlomi. Perhaps I am suffering from a bad case of http://xkcd.com/386/, but all I wanted to do by contacting you on Formspring (since you have no email listed) was to follow up on this post – privately(!) – and see if you still felt there was no problem. I don’t feel that was too unreasonable, although I again apologize for my phrasing.

Respectfully,

Zev

Comment by Zev Chonoles | February 27, 2011 | Reply