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4 More clarification

For (suitable) real- or complex-valued functions f and g on a (suitable) abelian group G, we have two bilinear operations: multiplication -

(f.g)(x) = f(x)g(x),

and convolution -

(f*g)(x) = ∫y+z=xf(y)g(z)

Both operations define commutative ring structures (possibly without identity) with the usual addition. (For that to make sense, we have to find a subset of functions that is closed under addition, multiplication, and convolution. If G is finite, this is not an issue, and if G is compact, we can consider infinitely differentiable functions, and if G is Rd, we can consider the Schwarz class of infinitely differentiable functions that decay at infinity faster than all polynomials, etc. As long as our class of functions doesn't satisfy any additional nontrivial algebraic identities, it doesn't matter what it is precisely.)

My question is simply: do these two commutative ring structures satisfy any additional nontrivial identities?

A "trivial" identity is just one that's a consequence of properties mentioned above: e. g., we have the identity

f*(g.h) = (h.g)*f,

but that follows from the fact that multiplication and convolution are separately commutative semigroup operations.

Edit: to clarify, an "algebraic identity" here must be of the form "A(f1, ... fn) = B(f1, ..., fn)," where A and B are composed of the following operations:

• addition
• negation
• additive identity (0)
• multiplication
• convolution

(Technically, a more correct phrasing would be "for all f1, ..., fn: A(f1, ... fn) = B(f1, ..., fn)," but the universal quantifier is always implied.) While it's true that the Fourier transform exchanges convolution and multiplication, that doesn't give valid identities unless you could somehow write the Fourier transform as a composition of the above operations, since I'm not giving you the Fourier transform as a primitive operation.

Edit 2: Apparently the above is still pretty confusing. This question is about identities in the sense of universal algebra. I think what I'm really asking for is the variety generated by the set of abelian groups endowed with the above five operations. Is it different from the variety of algebras with 5 operations (binary operations +, *, .; unary operation -; nullary operation 0) determined by identities saying that (+, -, 0, *) and (+, -, 0, .) are commutative ring structures?

3 Clarification

For (suitable) real- or complex-valued functions f and g on a (suitable) abelian group G, we have two bilinear operations: multiplication -

(f.g)(x) = f(x)g(x),

and convolution -

(f*g)(x) = ∫y+z=xf(y)g(z)

Both operations define commutative ring structures (possibly without identity) with the usual addition. (For that to make sense, we have to find a subset of functions that is closed under addition, multiplication, and convolution. If G is finite, this is not an issue, and if G is compact, we can consider infinitely differentiable functions, and if G is Rd, we can consider the Schwarz class of infinitely differentiable functions that decay at infinity faster than all polynomials, etc. As long as our class of functions doesn't satisfy any additional nontrivial algebraic identities, it doesn't matter what it is precisely.)

My question is simply: do these two commutative ring structures satisfy any additional nontrivial identities?

A "trivial" identity is just one that's a consequence of properties mentioned above: e. g., we have the identity

f*(g.h) = (h.g)*f,

but that follows from the fact that multiplication and convolution are separately commutative semigroup operations.

Edit: to clarify, an "algebraic identity" here must be of the form "A(f1, ... fn) = B(f1, ..., fn)," where A and B are composed of the following operations:

• addition
• negation
• additive identity (0)
• multiplication
• convolution

(Technically, a more correct phrasing would be "for all f1, ..., fn: A(f1, ... fn) = B(f1, ..., fn)," but the universal quantifier is always implied.) While it's true that the Fourier transform exchanges convolution and multiplication, that doesn't give valid identities unless you could somehow write the Fourier transform as a composition of the above operations, since I'm not giving you the Fourier transform as a primitive operation.

2 corrected the typo

For (suitable) real- or complex-valued functions f and g on a (suitable) abelian group G, we have two bilinear operations: multiplication -

(f.g)(x) = f(x)g(x),

and convolution -

(f*g)(x) = ∫y+z=xf(y)g(z)

Both operations define commutative ring structures (possibly without identity) with the usual addition. (For that to make sense, we have to find a subset of functions that is closed under addition, multiplication, and convolution. If G is finite, this is not an issue, and if G is compact, we can consider infinitely differentiable functions, and if G is Rd, we can consider the Schwarz class of infinitely differentiable functions that decay at infinity faster than all polynomials, etc. As long as our class of functions doesn't satisfy any additional nontrivial algebraic identities, it doesn't matter what it is precisely.)

My question is simply: do these two commutative ring structures satisfy any additional nontrivial identities?

A "trivial" identity is just one that's a consequence of properties mentioned above: e. g., we have the identity

f*(g.h) = (h.g)*h,h.g)*f,

but that follows from the fact that multiplication and convolution are separately commutative semigroup operations.

1