Let $A$ be a commutative ring and $M_i, i \in I$ be a infinite family of $A$-modules. Define their tensor product $\bigotimes_{i \in I} M_i$ to be a representing object of the functor of multilinear maps defined on $\prod_{i \in I} M_i$ (this exists by the usual construction). Thus there is a universal multilinear map $\otimes : \prod_{i \in I} M_i \to \bigotimes_{i \in I} M_i$. Some years ago, I wanted to examine this infinite tensor product, but in the literature I could not find anything going beyond some natural isomorphisms (e.g. associativity) or the submodule consisting of tensors which become eventually constant a specific element of $\prod_{i \in I} M_i$ which yields a colimit of finite tensor products (denoted $U_x$ below). In general, it seems to be quite hard to describe $\bigotimes_{i \in I} M_i$. For example, for a field $K$, $K \otimes_K \otimes_K ...$ has dimension $|K^*|^{\aleph_0}$ (see below) and you cannot write down a basis, which might be scary when you see it the first time. The point is that multilinear relations cannot be applied infinitely many times at once: For example in $K \otimes_K \otimes_K ...$, we have $x_1 \otimes x_2 \otimes ... = y_1 \otimes y_2 \otimes ...$ if and only if $x_i = y_i$ for almost all $i$ and for the rest we have $\prod_i x_i = \prod_i y_i$.

Before posing my question, I provide some results.

1.1. Assume that $M_i$ are torsionfree $A$-modules (meaning $am=0 \Rightarrow a=0 \vee m=0$). In this case, we may decompose $\bigotimes_{i \in I} M_i$ as follows: Define $X = \prod_{i \in I} M_i \setminus \{0\}$ and let $x \sim y \Leftrightarrow \{i : x_i \neq y_i\}$ finite. Then $\sim$ is an equivalence relation on $X$. Let $R$ be a set of representatives (this makes this description ugly!). Then there is a canonical map

$H : \text{Mult}(\prod_{i \in I} M_i,-) \to \prod_{x \in R} \lim_{E \subseteq I \text{~finite}} \text{Mult}^x(\prod_{i \in E} M_i,-)$,

where $\text{Mult}^x$ indicates that the transition maps of the limit are given by inserting the entries of $x$, and it is not hard to show that $H$ is bijective. Thus

$\bigotimes_{i \in I} M_i = \bigoplus_{x \in R} U_x$,

where $U_x = \cup_{E \subseteq I \text{~finite}} \bigotimes_{i \in E} M_i \otimes \otimes_{i \notin E} x_i$ is the colimit of the finite tensor products $\otimes_{i \in E} M_i$ (the transition maps given by tensoring with entries of $x$). The canonical maps $\otimes_{i \in E} M_i \to U_x$ don't have to be injective; at least when $A$ is a PID, this is the case. Remark that $U_x$ only depends on the equivalence class of $x$, so that the decomposition into the $U_x$ is canonical, whereas the representation of $U_x$ as direct limit (including the transition maps!) depends really on $x$.

1.2 If $M_i = A$ is an integral domain, we get $\bigotimes_{i \in I} A = \oplus_{x \in R} U_x$, where $U_x$ is the direct limit of copies $A_E$ of $A$, for every finite subset $E \subseteq I$, and transition maps $\prod_{i \in E' \setminus E} x_i : A_E \to A_{E'}$ for $E \subseteq E'$. $U_x$ is just the localization of $A$ at the $x_i$.

1.3 If $A$ is a field, and $M_i$ has basis $B_i$, then $B_x = \cup_{E \subseteq I} \bigotimes_{i \in E} B_i \otimes \otimes_{i \notin E} x_i$ is a basis of $U_x$ and thus $\cup_{x \in R} B_x$ is a basis of $\bigotimes_{i \in I} M_i$. According to this question, this has cardinality $\max(|X|,|I|,\max_i(\dim(M_i)))$.

1.4 If $A_i$ are $A$-algebras, then $\bigotimes_{i \in I} A_i$ is a $A$-algebra. If the $A_i$ are integral domains, then it is a graded algebra by the monoid $X/\sim$ with components $U_x$.

If $A=A_i=K$ is a field with $U=K^x$, then there is a vector space isomorphism between $\bigotimes_{i \in I} K$ and the group algebra $K[U^I / U^{(I)}]$. A sufficient, not neccessary, condition for the existence of a $K$-algebra isomorphism is that $U^{(I)}$ is a direct summand of $U^I$, which is quite rare (see this question). Nevertheless, we can ask if these $K$-algebras isomorphic. In some sense I have proven this already locally (subalgebras given by finitely generated subgroups of the group $U^I / U^{(I)}$ are isomorphic, in a terribly uncanonical way). Many questions I'm currently posing here are addressed to this problem.

2.1 What about interchanging tensor product with duals? Let $(V_i)_{i \in I}$ be a family of vector spaces over a field $K$. For elements $\lambda_i \in K$, define their infinite product $\prod_{i \in I} \lambda_i$ to be the usual product if $\lambda_i=1$ for almost all $i$, and otherwise to be $0$. This yields a multilinear map $\prod : K^I \to K$ and thus a linear map $\delta : \bigotimes_{i \in I} V_i^* \to (\bigotimes_{i \in I} V_i)^*, \otimes_i f_i \mapsto (\otimes_i x_i \mapsto \prod_{i \in I} f_i(x_i)).$ Then it can be shown that $\delta$ is injective, but the proof is pretty fiddly.

2.2 Let $W_i$ be another family of vector spaces over a field $K$. Then there is a canonical map $\alpha : \bigotimes_{i \in I} \operatorname{Hom}(V_i,W_i) \to \operatorname{Hom}(\bigotimes_{i \in I} V_i, \bigotimes_{i \in I} W_i).$ Is $\alpha$ injective? This is known when $I$ is finite.

3 What about other properties of finite tensor products, do they generalize? For example let $J_i, i \in I$ be a family of index sets and $M_{i,j}$ be a $A$-module where $i \in I, j \in J_i$. Then there is a canonical homomorphism $\delta : \bigoplus_{k \in \prod_{i \in I} J_i} \bigotimes_{i \in I} M_{i,k(i)} \to \bigotimes_{i \in I} \oplus_{j \in J_i} M_{i,j}$. It can be shown that $\delta$ is injective, but is it also bijective (as in the finite case)?

4 The description of the tensor product given in 1.1 depends on a set of representatives and is not handy when you want to prove something. Are there better descriptions?

Remark that in this question I'm not interested in infinite tensor products defined in functional analysis or just colimits of finite ones. I'm interested in the tensor product defined above (which probably every mathematician regards as "the wrong one"). Any hints about their structure or literature about it are appreciated.

  • $\begingroup$ I've addded a "category theory" tag since I think some of the background issues in the question might be related to things categorists have looked at. If people disagree, please feel free to retag $\endgroup$
    – Yemon Choi
    Jan 14, 2010 at 20:57
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    $\begingroup$ of course the direct product, otherwise every multilinear map would be trivial (with the usual definition). mathematics becomes interesting as soon as you think about something which does not follow trivially from category theory ... ;-). of course I already thought of alternative definitions (and the results above are only a selection). $\endgroup$ Jul 8, 2010 at 13:40
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    $\begingroup$ So basically I think it is a bad idea to impose only finitely many multilinear relations even for infinitely many factors, but every attempt to impose some sort of infinitely many multilinear relations makes every tensor product $0$. $\endgroup$ Jul 19, 2010 at 8:09
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    $\begingroup$ I had read (student time) about infinite multiliear(and infinite tensoring) in the Claude Chevalley "Fundamental concepts of algebra". May be you just know this reference, $\endgroup$ Dec 15, 2011 at 8:27
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    $\begingroup$ Read more carefully, this is not what is written there. It is said that $\bigotimes_{i \in I} M_i$ exists, and the same proof as in the finite case works. Take the free module on $\prod_{i \in I} |M_i|$ and quotient out the multilinear relations. $\endgroup$ Feb 4, 2020 at 23:04

1 Answer 1


Coincidentally, I started a work in 2009 concerning infinite tensor products of complex vector spaces, $^*$-algebras and Hilbert spaces. The resulting article has just been accepted for publication and I put it in the arxiv: https://arxiv.org/abs/1112.3128. In the article, I got a similar result as your 1.1 (for vector spaces) and obtained a positive answer to your question 2.2 (when $K$ is the complex filed).

  • $\begingroup$ Thanks! More generally the proof of 2.2. works for infinite fields. $\endgroup$ Jan 3, 2012 at 10:04
  • $\begingroup$ If the paper has been accepted, can you put the publication reference on arxiv ? $\endgroup$ Mar 13 at 5:47

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