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Given a noncommutative ring $R$, and two (left) $R$-modules $M$ and $N$, how does one define a left action on the the vector space tensor product $M \otimes N$? Multiplying on just the first factor of the tensor product seems a little unnatrual, but I can't see what else to do.

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You usually do this when $R$ also has a coalgebra structure $R \to R \otimes R$. –  Angelo Sep 22 '12 at 14:30
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What do you mean by the tensor product of two left R-modules? Normally when R is noncommutative, you need a left R-module and a right R-module, and then the tensor product doesn't have an R-structure. –  anon Sep 22 '12 at 14:30
    
With a definition like $r \triangleright (m \otimes n) = \sum_i r_i m \otimes r'_i n$, where $\Delta(r) = \sum_i r_i \otimes r_i'$? –  Ago Szekeres Sep 22 '12 at 14:39
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I'd agree with Angelo, you need an algebra map $R \to R\otimes R$. Without more information on $R$ the only two such maps are $r \mapsto r \otimes 1$ and $r \mapsto 1 \otimes r$, which make $M \otimes N$ into a left module by multiplying the $M$ or $N$ factor, respectively. This usually isn't a reasonable thing to do, I think. –  Peter Samuelson Sep 22 '12 at 16:08
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@Ago: because both constructions ignore the fact that one of $M$ and $N$ is also a module. What do you mean by "vector space tensor product" if $R$ is not an algebra over a field? –  Qiaochu Yuan Sep 22 '12 at 18:47

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up vote 10 down vote accepted

Let $R, S$ be two (unital and associative to be safe) algebras over a commutative ring $k$ and let $M, N$ be respectively a left $R$-module and a left $S$-module. Then we can define the tensor product $M \otimes_k N$ by the usual universal property, and it is naturally a left $R \otimes_k S$-module by functoriality. If $M, N$ are both left $R$-modules, then $M \otimes_k N$ is a left $R \otimes_k R$-module, so defining a left $R$-module structure on the tensor product which is compatible with its $k$-module structure is tantamount to giving a morphism of $k$-algebras

$$\Delta : R \to R \otimes_k R.$$

Note that any such morphism gives a left $R$-module structure to the tensor product of two left $R$-modules, but it won't behave the way you expect it to with respect to multiple tensor products unless $\Delta$ is coassociative and cocommutative. I will be ignoring this.

Now, the ring $R \otimes_k R$, by its universal property, naturally admits two morphisms $R \to R \otimes_k R$, namely the two inclusions $i_1, i_2 : R \to R \otimes_k R$, which correspond to the two $R$-module structures described by Peter Samuelson in the comments. If you don't want to just provide $\Delta$ as extra data, then your question can be reinterpreted as follows:

What other natural morphisms $R \to R \otimes_k R$ are there?

Suppose that $\Delta_R : R \to R \otimes_k R$ is a family of morphisms which is natural in $R$; that is, it defines a natural transformation. Applying the forgetful functor to $\text{Set}$, we get a natural transformation from the forgetful functor $U : k\text{-Alg} \to \text{Set}$ to the functor

$$U^2 : k\text{-Alg} \ni R \mapsto U(R \otimes_k R) \in \text{Set}.$$

The functor $U$ is representable by the $k$-algebra $k[x]$, so by the Yoneda lemma, natural transformations $U \to U^2$ can be naturally identified with elements of

$$U^2(k[x]) \cong U(k[x, y]).$$

More concretely, any polynomial in two variables $f(x, y) = \sum f_{ij} x^i y^j \in k[x, y]$ over $k$ induces a natural map of sets

$$R \ni r \mapsto \sum f_{ij} r^i \otimes r^j \in R \otimes_k R$$

and these are the only such natural maps. Now the question reduces to the following:

Which of the maps $r \mapsto \sum f_{ij} r^i \otimes r^j$ are always morphisms of $k$-algebras?

The short answer is that it depends a lot on $k$. For starters, letting $R = k[x], r = x, R \otimes_k R = k[x, y]$ and requiring compatibility with scalar multiplication gives

$$cr \mapsto c f(x, y) = f(cx, cy) \in k[x, y]$$

for all $c \in k$. Comparing coefficients of $x^i y^j$ gives

$$(c - c^{i+j}) f_{ij} = 0$$

for all $i, j$ and for all $c$. This suggests the possibility of other natural morphisms $R \to R \otimes_k R$ in characteristic $p$. In particular, if $k = \mathbb{F}_p$ then we have $c^{p^r} = c$ for all $r \ge 0$, hence $f$ can have components of degree $p^r$ and some of these give rise to genuine natural morphisms such as

$$R \ni r \mapsto r^p \otimes 1 \in R \otimes_k R.$$

Assume for simplicity that $k$ is an integral domain of characteristic $0$ (these hypotheses can probably be considerably weakened). Then the above condition implies that $f$ is homogeneous and linear, so our desired morphism can only have the form

$$R \ni r \mapsto f_{10} r \otimes 1 + f_{01} 1 \otimes r \in R \otimes_k R.$$

Compatibility with multiplication now requires (taking $R = k[x, y], R \otimes_k R = k[x, y, z, w]$)

$$xy \mapsto f_{10} xy + f_{01} zw = (f_{10} x + f_{01} z)(f_{10} y + f_{01} w)$$

and comparing coefficients gives

$$f_{10}^2 = f_{10}, f_{01}^2 = f_{01}, f_{10} f_{01} = 0.$$

Our hypotheses then imply that exactly one of $f_{10}$ or $f_{01}$ can be equal to $1$ and the other must be equal to $0$. (If $k$ is not an integral domain then other things can happen; in general a choice of $f_{01}, f_{10}$ above is equivalent to a direct product decomposition $k = k_1 \times k_2$ with respect to which $f_{10} = (1, 0), f_{01} = (0, 1)$.) In other words, when $k$ is an integral domain of characteristic $0$ then the only natural morphisms $R \to R \otimes_k R$ are

$$R \ni r \mapsto r \otimes 1 \in R \otimes_k R$$

resp.

$$R \ni r \mapsto 1 \otimes r \in R \otimes_k R$$

and the corresponding natural left $R$-module structures on $M \otimes_k N$ are given by multiplication on the first resp. the second factor. In particular,

If $k$ is a field of characteristic $0$, then the only natural morphisms $R \to R \otimes_k R$ are the two obvious ones.

Group algebras are highly misleading here: any group algebra $k[G]$ (more generally any monoid algebra $k[M]$) is naturally equipped with a comultiplication given by extending

$$\Delta : k[G] \ni g \mapsto g \otimes g \in k[G] \otimes_k k[G]$$

and $\Delta$ induces the usual tensor product of representations of $G$ over $k$. It is worth mentioning that this map is not defined using any of the structure maps of the group $G$; more generally, for any set $S$ there is a canonical diagonal map $\Delta : S \to S \times S$ given by sending $s$ to $(s, s)$, and this induces a canonical coalgebra structure on the free module $k[S]$ given by extending

$$k[S] \ni s \mapsto s \otimes s \in k[S] \otimes_k k[S].$$

But a generic $k$-algebra is not a free vector space over any distinguished set.

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Well put: explaining, in effect, what "reasonable" should be, which was part of the question's "problem", and then decisively answering. –  paul garrett Sep 22 '12 at 21:56
    
It's nice to see that the intuition that the two obvious morphisms $R \to R\otimes_k R$ are the only "natural" ones (in characteristic 0) can be formalized by removing the quotation marks :) –  Peter Samuelson Sep 23 '12 at 4:35

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