Re: JSH: So why tautological spaces?

Based on your original post, this thread was and still is about
"tautological spaces", and how you claim them to be a new mathematical
tool you invented and that maths people are trying to suppress via
journal murder and whatnot. In reply to that claim I pointed out that
these "tautological spaces" are not new at all, and are simply an
example of the more general machinery of abstract algebra,
specifically a quotient algebra.

A claim I had no interest in evaluating.  Usenet is an arena where
people routinely lie.

I'm not going to go on some wild goose chase after papers which you
claim are in French anyway just because you said so, as it was simpler
to ask you to demonstrate.

A mathematical demonstration cuts through the b.s. as anyone can SAY
something is not new.

I could SAY that Einstein was an alien from the planet Kalamazoo.  But
that doesn't make it true.

(Hey that all rhymed.)

In response to this, instead of addressing my rebuttal of your claim

I rebut your rebuttal.  And claim instead that you just babbled some
stuff.

Now that rebut of your rebuttal should be rebut worthy for anyone who
is noting your refusal to just DO THE MATH.

You could just be some talkative nincompoop like so many Usenet
posters.

You may not know ANY math.  What math DO you know by the way?  What is
your specialty?

By answering at least MAYBE readers can get some sense of whether or
not you know anything about mathematics at all.

you instead challenged me to use this abstract algebra to derive a
general reduction form for QDE's. I am curious as to what point you

Something I've done.  And I've given the MATHEMATICS.  So people can
check me.

You have talked and made verbal claims, and refused to do a
demonstration.

were trying to make by making this challenge. Why would I do this
talk? What would this show? It is going to change the fact that your

A demonstration would show that you're not lying.  Mathematics either
works or it does not.

"tautological spaces" are just quotient algebras and are well known
and used in maths already?

Then it should be easy to demonstrate.

Ok.  I'll make it easier.  You don't even have to complete the
reduction.  Just do the setup.  The initial work towards solving the
problem given, and then you can even TALK out how it would be
completed giving you yet another opportunity to babble.

The check here is a reality check: Usenet is a free speech zone, and
some people use their free speech to lie or tell falsehoods (assuming
that maybe they believe what they're saying).

The check for such behavior with mathematics is a request for a
mathematical demonstration.

Refusal to do something you claim is easy and well-known does not look
good.

Less talk--more math action!

James Harris


The first thing I will explain is that I provided links and references
in my posts because if something is written up in the literature (and
especially those things written up nicely) there is not much point
redoing it - just point to the reference. This is one of the key
aspects of human intellectual study, we build on existing work.

This is particularly relevant for your continued insistence on
deriving this QDE simplification - IT IS RIGHT HERE

http://www.alpertron.com.ar/METHODS.HTM

Bang. Done. Alpern has a beautiful, very clear and well written
writeup of this exact problem. So why would I bother repeating that
here? Moreover, as I maintained before and as the alpern site
demonstrates, you dont need to quotient the algebra of polynomials in
order to get the simplification - you just obtain it on the nose.

I will address your brash demands for a demonstration of the process
of quotienting the polynomial algebra later in this post but first I
will answer your demands as to my maths specialization.

I work in the field of non-commutative geometry, studying the geometry
of non-commutative C*-algebras using spectral triples, machinery
developed by Connes in the 1980's. This entails a mix of analysis and
algebra. The analytic work looks at building the equivalent of an
elliptic operator on non-commutative manifolds. This can then be used
to study geometric information about the C*-algebra as the elliptic
operator defines a natural element in the cyclic cohomology of the
algebra, via commutators of the operator with algebra elements.

So, algebra is a pretty fundamental tool in what I do. Indeed algebra
quotients very often arise in this situation as we often want to
compute the K-theory and K-homology of our algebra. We want to compute
these because one of the interesting geometric invariants in this
picture is the pairing between K-theory and homology. Ideals and
quotients by ideals give rise to short exact sequences in an immediate
way, which in turn give rise to 6-term exact sequences in KK-theory.
These are a very common, and useful, way of computing the KK-theory of
a C*-algebra, and this is normally where I use quotients in my work.

So, you wanted a demonstration where your "tautological spaces" come
from? OK, here we go. First we let A = C[x, y, z], the unital algebra
of polynomials in three commutating generators x, y and z with complex
coefficients. Feel free to take real coefficients as it changes
nothing. We trivially observe that this is indeed an algebra as
multiplication and addition of polynomials easily gives polynomials
again. We are not interested in equipping the algebra with a topology
at this stage.

Now we define an equivalence relation on the algebra A. For p, q in A
we say

p ~ q if there exists an element a in A such that
p = q + a(x + y + vz)

where v is a fixed complex coefficient. Now we check that this is
indeed an equivalence relation

Reflexive: Since p = p then p ~ p (taking a = 0 in A) for all p in A

Symmetric: Suppose p ~ q, that is p = q + a(x + y + vz) for some a in
A. Then q = p - a(x + y + vz), and since -a is in A we have that q ~ p

Transitive: Suppose p ~ q and q ~ r, that is p = q + a(x + y + vz) and
q = r + b(x + y + vz) for some a, b in A. Then p = r + (a + b)(x + y +
vz), and since a, b in A we have a+b in A, so p ~ r.

Therefore ~ is an equivalence relation on A. Now we consider the set
formed by the equivalence class of 0 (it turns out to be an ideal).
That is, let

I = {a in A | a ~ 0}

Clearly I = {a(x + y + vz) | for all a in A}, which we may write as I
= (x + y + vz)A = A(x + y + vz) since A is commutative. Now we show
that I is an ideal of A. First we observe that I is closed under
addition

a(x + y + vz) + b(x + y + vz) = (a+b)(x + y + vz)

which is in I as (a+b) is in A. Next we observe that for any a in A
and i in I, ai = ia is in I. This is because i = b(x + y + vz) for
some b in A, so ai = ab(x + y + vz) which is in I, as ab is in A.

So, we have an ideal generated by the equivalence class of 0 under the
equivalence relation ~. Now we take the quotient of A by I, which we
denote (using the standard notation) A/I. The quotient map Q : A -> A/
I is an algebra homomorphism, meaning

Q(p + q) = Q(p) + Q(q)
Q(pq) = Q(p)Q(q)

The quotient map is given by the left (or right, this is commutative)
cosets of I, so

Q(p) = p + I

Now we can see why we needed an ideal to quotient by in order to get a
quotient algebra, not just a quotient (additive) group. This is
because

Q(p)Q(q) = (p + I)(q + I)
= pq + pI + qI + I

So if pI did not land in the ideal for all p in A then the quotient
map would not be a homomorphism under multiplication. Since we do have
an ideal, we get

Q(p)Q(q) = pq + I + I + I = pq + I = Q(pq)

So, now it is clear that elements in the quotient algebra are of the
form

p + a(x + y + vz)

for some a in A. What more is there to say? This is just a nice, basic
example of quotienting an algebra by an ideal, case closed. What did
we gain by doing it this way instead of JSH's way? Well apart from
clarity we gained the important information that multiplication is
well defined in the quotient algebra, and we have a MATHEMATICAL PROOF
that this is true, which JSH never supplied. He simply appealed to
intuition and common sense with a few test case examples.

JSH, feel free (as long as you credit me at the top) to copy and paste
this onto the "tautological space" page of your blog, as this is a
general definition of your "tautological spaces". Also, if you keep
talking about them can you please call them algebras, as that is what
they are.

DH
.

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