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From: Jay R. Yablon on 23 Feb 2010 16:18
Please take a look at http://en.wikipedia.org/wiki/Lie_product_formula.

Would this formula, which is stated to apply to NxN matrices, also apply
to an infinite-dimensioned operator / matrix? Why or why not?

Thanks,

Jay
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Email: jyablon(a)nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
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From: Stephen Montgomery-Smith on 24 Feb 2010 22:36
Jay R. Yablon wrote:
> Please take a look at http://en.wikipedia.org/wiki/Lie_product_formula.
>
> Would this formula, which is stated to apply to NxN matrices, also apply
> to an infinite-dimensioned operator / matrix? Why or why not?

I remember reading a book on semigroups of linear operators, where they
proved that given an unbounded operator A, with certain properties on
the resolvent, that there exists a semigroup of bounded operators G(t)
(t>=0) such that G(s)G(t)=G(s+t) (that is what semigroup means), G(t)
converges strongly to the identity as t->0, and (G(t)-I)/t converges
strongly on the domain of A.

I'm sorry I don't remember all the hypotheses, not do I remember the
author of the book. But I do remember that the proof used the Lie
produce formula.

This leads me to believe the formula must be true for bounded operators.

Maybe the book by Jan van Neerven http://fa.its.tudelft.nl/~neerven/
http://www.ams.org/mathscinet-getitem?mr=98d:47001 will have something
about this.

Stephen
From: Robert Israel on 25 Feb 2010 16:59
Stephen Montgomery-Smith <stephen(a)math.missouri.edu> writes:

> Jay R. Yablon wrote:
> > Please take a look at http://en.wikipedia.org/wiki/Lie_product_formula.
> >
> > Would this formula, which is stated to apply to NxN matrices, also apply
> > to an infinite-dimensioned operator / matrix? Why or why not?
>
> I remember reading a book on semigroups of linear operators, where they
> proved that given an unbounded operator A, with certain properties on
> the resolvent, that there exists a semigroup of bounded operators G(t)
> (t>=0) such that G(s)G(t)=G(s+t) (that is what semigroup means), G(t)
> converges strongly to the identity as t->0, and (G(t)-I)/t converges
> strongly on the domain of A.
>
> I'm sorry I don't remember all the hypotheses, not do I remember the
> author of the book. But I do remember that the proof used the Lie
> produce formula.
>
> This leads me to believe the formula must be true for bounded operators.

Yes, although it's mainly interesting for semigroups generated by unbounded
operators. See e.g. <http://eom.springer.de/T/t094340.htm>
or Reed and Simon, Methods of Modern Mathematical Physics I. Functional
Analysis, Theorem VIII.31.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Jay R. Yablon on 25 Feb 2010 19:19

"Robert Israel" <israel(a)math.MyUniversitysInitials.ca> wrote in message
news:rbisrael.20100225213953$2948(a)news.acm.uiuc.edu...
> Stephen Montgomery-Smith <stephen(a)math.missouri.edu> writes:
>
>> Jay R. Yablon wrote:
>> > Please take a look at
>> > http://en.wikipedia.org/wiki/Lie_product_formula.
>> >
>> > Would this formula, which is stated to apply to NxN matrices, also
>> > apply
>> > to an infinite-dimensioned operator / matrix? Why or why not?
>>
>> I remember reading a book on semigroups of linear operators, where
>> they
>> proved that given an unbounded operator A, with certain properties on
>> the resolvent, that there exists a semigroup of bounded operators
>> G(t)
>> (t>=0) such that G(s)G(t)=G(s+t) (that is what semigroup means), G(t)
>> converges strongly to the identity as t->0, and (G(t)-I)/t converges
>> strongly on the domain of A.
>>
>> I'm sorry I don't remember all the hypotheses, not do I remember the
>> author of the book. But I do remember that the proof used the Lie
>> produce formula.
>>
>> This leads me to believe the formula must be true for bounded
>> operators.
>
> Yes, although it's mainly interesting for semigroups generated by
> unbounded
> operators. See e.g. <http://eom.springer.de/T/t094340.htm>
> or Reed and Simon, Methods of Modern Mathematical Physics I.
> Functional
> Analysis, Theorem VIII.31.
> --
> Robert Israel israel(a)math.MyUniversitysInitials.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada

Thank you, Robert. I found the Springer link, but the book is very
helpful and right on point filling in a lot more detail. Jay

From: Don Stockbauer on 25 Feb 2010 21:05
On Feb 23, 3:18 pm, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote:
> Please take a look athttp://en.wikipedia.org/wiki/Lie_product_formula.
>
> Would this formula, which is stated to apply to NxN matrices, also apply
> to an infinite-dimensioned operator / matrix?  Why or why not?

You failed to specify whether you mean the potential or the actualized
infinity.
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