From: johnae@ix.netcom.com (John Eisenberg)
Newsgroups: sci.math,sci.econ,scri.math.research
Subject: Smooth associative operations question
Date: Wed, 22 Oct 1997 21:43:22 GMT
I am posting this for a friend of mine. Please respond to him at
marschak@socrates.berkeley.edu
Suppose that * is an associative operation on R such that the function
F(x_1,...,x_n) = x_1 * x_2 * ... * x_n is smooth then what can we
say about * ?
Specifically, must we have
a*b = g(a) + g(b) or
a*b = g(a)*g(b) for some smooth g : R --> R ?
Thank you.
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From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math,sci.econ
Subject: Re: A math/econ question
Date: 23 Oct 1997 15:48:44 GMT
In article <344bb3a4.229507@nntp.ix.netcom.com>,
John Eisenberg wrote:
>Let F be a smooth (C-infinity) real valued function on
>T = T_1 X T_2 X ... X T_u where the T_i are open subsets of R.
>
>Suppose that F(t_1,...,t_u) = t_1 * t_2 * ... * t_u where "*"
>is an associative operation.
>Then what can be said about "*"? Must it be of the form
>
>a*b = g(a) + g(b) for g smooth?
Not at all. At the very least, you should adjust this proposal to make
'*' really be associative, which your '*' isn't for general g (e.g.
g(x)=x^2). A natural alternative is to ask if a*b = g^(-1) ( g(a) + g(b) ),
for some smooth invertible g. These '*' _are_ all associative, for
then a*b*c would be g^(-1) ( g(a) + g(b) + g(c) ) irrespective of the
placement of parentheses.
But then this question seems to ask, is there a smooth semigroup operation '*'
on the real line which is not, up to conjugacy by diffeomorphisms, just
addition? Here the answer is again 'no', since there are
semigroup structures which are not group structures. Taking a*b to be
identically zero gives one example. (So does a*b = |a|+|b|, although this
product map isn't smooth.)
Finally, note that the question you posed isn't quite well defined. Are we
assuming '*' is defined on all of R or only on some subset (say, the
union of the T_i's) ? It makes a difference; in the one case, we can
take g(x) = log(x), so that a*b = exp(log(a)+log(b)) is ordinary
multiplication; but you cannot write ordinary multiplication as a
conjugate of addition in this way if you insist that it be defined for all
of R (note log(x) is undefined for x<=0).
dave
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From: synth23@ix.netcom.com (C. M.)
Newsgroups: sci.math
Subject: Associative operations and smooth functions question
Date: Sat, 25 Oct 1997 23:15:08 GMT
I'm posting this for a friend of mine. Please email responses to him
at marschak@socrates.berkeley.edu
Let Q be an open subset of R, F: Q X Q X Q --> R smooth. Also, assume
F is symmetric.
Let A and G be smooth real-valued functions such that
F(x,y,z) = G(A(x,y),z).
Can we conclude that there exists an associative operation * and a
smooth real-valued function, g, such that:
F(x,y,z) = g(x) * g(y) * g(z)?
Thanks very much.
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From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math,sci.econ,scri.math.research
Subject: Re: Smooth associative operations question
Date: 27 Oct 1997 17:21:17 GMT
In article <344e7198.14850504@nntp.ix.netcom.com>,
John Eisenberg wrote:
>Suppose that * is an associative operation on R such that the function
>
>F(x_1,...,x_n) = x_1 * x_2 * ... * x_n is smooth then what can we
>say about * ?
>
>Specifically, must we have
>
>a*b = g(a) + g(b) or
>a*b = g(a)*g(b) for some smooth g : R --> R ?
No. As I remarked in a previous post, one must certainly allow for
isomorphisms (e.g. a*b = h^(-1)(h(a)+h(b)) with, say, h(x)=x+exp(-x^2)
is well-defined and smooth, but it's really just "+"). In addition, I
should have mentioned projections such as x*y=y, obviously smooth
and associative.
I'm beginning to think less is known about associative, continuous
binary operations on R than I had previously suspected. I believe that
the assumption of differentiability will help rule out most of the
anomolous products, but the situation is not nearly as clear for
differentiable semigroups as for groups. What appears to be the right
paper establishing the role of differentiability is
Graham, George E. Differentiable semigroups. Recent
developments in the algebraic, analytical, and topological theory of
semigroups (Oberwolfach, 1981), 57--127, Lecture Notes in Math., 998,
Springer, Berlin-New York, 1983
Without the differentiability hypothesis, it's possible to pin down the
structure somewhat but only with some additional hypotheses about '*'.
Perhaps (given your conjectured form for '*') some of those hypotheses
are appropriate for you: do you think '*' is commutative? Is it
cancellative (x*y=x*z implies y=z)? Is there an identity (i*x=x for all x)
or an idempotent (e*e=e) or a zero (z*x=z for all x)?
A possible general reference is Paul S. Mostert, The structure of topological
semigroups---revisited. Bull. Amer. Math. Soc. 72 1966 601-618. Here are
some other papers I noticed which may be appropriate. (I haven't read them.)
J G Horne Jr,"Multiplications on the line", Proc Amer Math Soc 9 (1958) 791-795
classifies the ones which coincide with usual multiplication on [0, oo)
(and having 0*x=0 and 1*x=x even for negative x). They are isomorphic
either to
x*y=y*x=xy if x >=0 or y >=0, x*y=y*x=-xy otherwise
or else for some alpha,
x*y= xy if y >=0
x*y= x^alpha y if y < 0 and x >=0
x*y= 0 if y<0 and x < 0.
R J Koch and A D Wallace, Admissibility of semigroup structures on continua,
Trans Amer Math Soc 88 (1958) 277-287: if I read the review correctly, this
work implies: if the set of products x*y is all of R, then either
'*' is isomorphic either to addition or left- or right-projection.
Paul S Mostert and Allen L Shields, Semigroups with identity on a manifold
Trans Amer Math Soc 91 (1959) 380-389. If '*' has an identity and no other
idempotents, then it makes a _group_ out of R, that is, '*' is
isomorphic to addition. These authors considered semigroup structures on
intervals too; a typical paper is in Ann Math 65 (1957) 117-143, and
shows the semigroups are essentially ordinary multiplication on [0,1],
the product x*y=max(xy, 1/2) on [1/2,1], and the product x*y=min(x,y) on [0,1].
The real line has an ordering ("<"); do you think '*' preserves it?
If so, consider A H Clifford, Totally ordered commutative semigroups,
Bull Amer Math Soc 64 (1958) 305-316.
dave
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