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I am trying to figure out what the correct notation for a multiplicative algebra should be. I've seen a bunch of weird and conflicting ways for writing these things in the literature, but they are all either too specialized or bring in extra baggage.

To be more specific, given a ring R and an R-module, M, with a finite basis B, we can identify the elements of M with functions of the form $f, g : B \to R$. Then we construct the multiplicative algebra on M according to the following rule:

$$ (f \, g)(x) = f(x) g(x) $$

My question is simple: What is the standard notation for this construction? I've seen a bunch of weird forms. For example, in the case where R is the field of complex numbers, we can write something like [; C^k(B) ;] for this algebra. Or if we are dealing with matrices, we can talk about Hadamard products.

But can we write something simpler? I would really just like to do something along the lines of say; define R(B) to be this multiplicative algebra and be done with it. The reason for this is that if we are working with something like Pontryagin duality or representation theory, the set B may be a group/monoid/whatever, and there could be multiple algebras we defined over it.

To illustrate why this is an issue, let $B = \mathbb R$ be the group of reals under addition, then we can define both a convolution algebra, $R[ \mathbb R ]$ and a multiplicative algebra $ R( \mathbb R )$ which are identified with the same R-module, but have totally different structures. (Though they are related by Pontryagin duality in the case where $R = \mathbb C$.)

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Twice during the reading of your question I thought "oh he just wants ___, oh wait never mind... he doesn't mean that..." by the end, I have no answer... –  BBischof Aug 9 '10 at 22:58
    
Hmm... Is it maybe too ambiguous? Another place where this also gets used is in the Banach algebra of multipliers, where I have seen people write $M(G)$ for locally compact abelian groups. However this fixes the underlying ring to be $\mathbb C$ and brings in a whole bunch of extra assumptions I really don't need. –  Mikola Aug 9 '10 at 23:20
    
And actually I just thought of another situation which is even worse. Consider the module of n x n matrices. Then there are at least 3 (!!!) distinct algebras: 1. The matrix algebra, with product given by matrix multiplication. 2. The multiplicative algebra, with the Hadamard product. 3. A Lie algebra, with the Lie bracket as the underlying operator. –  Mikola Aug 9 '10 at 23:24
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Dear Mikola, I would suggest that you clarify your question, perhaps both with an example and by adding more precision. For example, if $M$ assumed to be free of finite rank; otherwise the identification of elements of $M$ with functions on the basis doesn't seem to be well-defined. –  Matt E Aug 10 '10 at 0:21
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Dear Mikola, If you're still reading: my confusion stemmed from various parts of the question --- e.g. you said that you care about the case $B$ is finite, but then gave the example $B = \mathbb R$ as the end, and you wrote about an $R$-module with a basis $B$, but didn't specifically say that the module is free, which made me unsure if this is what you meant. –  Matt E Jan 22 '11 at 15:31

1 Answer 1

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If $B$ is any set (finite or not), and $R$ is a ring, then I would write $\prod_{x \in B} R$ to denote the product of a number of copies of $R$ indexed by $B$ --- this product is again a ring (with the component-wise operations). This is precisely the set of functions $f: B \to R$ with the operations defined pointwise.

I guess one could also write $R^B$ (set-theory notation for the functions from $B$ to $R$), but in the parts of algebra that I commonly read (commutative algebra, algebraic and arithmetic geometry) the product notation of my first paragraph is standard, and the exponential notation is not.

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Well, it has been almost a year since I asked this question and I have maybe learned a bit since then. I now would tend to agree that it is better to explicitly write out the product, since that makes the most sense for finite groups and generalizes readily to the non-commutative case. –  Mikola Jun 7 '11 at 19:12

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