Convolutions and transforms

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Many things you're used to in the study of function spaces are
operators:
- (linear!) endomorphisms of the codomain
  e.g., multiplication by scalar, $T (f_x,f_y)=(f_y,f_x)$
- group actions on the domain
  (e.g., $Tf(x)=f(-x)$; the odd/even bit)
- multiplication by a function
- convolution by a function
- derivative (in one-dimension),
  differentiation with respect to a vector field
- transforms (esp. Fourier)

in any ring (or even more generally) we have a bracket,
so for operators we have the composition bracket

(e.g., Nigel's $[\partial/\partial x, x]=\Id$ as the defining feature
of a non-commutative geometry?)
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an important class of operators are those that commute with the shift,
and are called "shift-invariant operators" -- a key example is the
derivative. Apparently all shift-invariant operators commute.
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A nice kind of transform is
$T_K f(x) = \int K(x,y)f(y) \dy$
where $K$ is called the \emph{kernel}; this is called an "integral
operator".
(these are used a lot in physics and analysis)
your key examples are the Fourier transform and the Laplace transform.

Convolutions are transforms of the kind
$T_g f(x) = \int f(y)g(x-y) \dy$
i.e., where $K(x,y)=g(x-y)$: a very simple kind of kernel.

The fourier transform is when
$K(n,x) = e^{-inx}$
(or $e^{ikx}$, or $e^{2\pi i kx}$, or maybe with a
$\frac{1}{2\pi}$ or $\frac{1}{\sqrt{2\pi}}$ in front;
conventions vary widely)
(this is the most mathematically meaningful of the lot, coming from
representation theory and all, but others give nice forms for
various purposes)
(for inverse fourier transform, take the conjugate, again perhaps with
a constant around).

Note that this is of the form $K(x,y)=g(xy)$, but is not of the form
$g(x-y)$ -- it's not a convolution, but it's still a very nice form.

The Laplace transform is when
$K(\alpha,x) = e^{-\alpha x}$ (and we generally take it on only one
side: $\cL[f](s) = \int_0^\infty f(x)e^{-sx}\dx$)
An important property is that it takes constant coefficient
differential operators (i.e., polynomials in the $\partial/\partial
x_i$ with constant coefficients) to polynomials.
Hence to solve a constant coefficient differential equation,
you can take the Laplace transform, solve the algebra, and then take
the inverse Laplace transform to get your solution;
this is called (generally) the Heaviside (he of the step function)
calculus.

This allows you to solve linear -inhomogenous- differential equations:
$p(D)f(x) = g(x)$
(solve for $f(x)$, where $p(D)$ is a polynomial in the derivative).
homogeneous linear differential eqns are easy linear algebra.

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note that the Gamma function
$\Gamma(z)= \int_0^\infty t^{z-1} e^{-t} \dt$
is the transform of $e^{-t}$ with respect to the kernel
$K(z,t)=t^{z-1}$, which is already a bit weird, but makes it seems
kinda natural...kinda.
I don't know if this transform has a name or geometric interpretation.
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transforms that are convolutions:
the identity is convolution with $\delta$,
while the derivative is convolution with $\delta'$!
(and shift by $-a$ is convolution with $\delta_a$)
(these are convolutions if you allow distributions, otherwise
 they're just abstract operators)
The Hilbert transform is convolution with $-1/(\pi x)$;
  dunno how it's applicable.

associativity of convolution is obvious, viewed as multiplication in
group algebra (as $\bR$ and $\bC$ are associative)
in particular, this means that if you apply successive transforms
(that are convolutions!), it's just applying the convolution of them,
i.e., here composition of operators is convolution of functions.

convolution with $e^{-x^2}$ is a global ``smoothing'' operator;
convolution with a bump function is a local smoothing
(in particular, if you convolve a continuous function with one of the
above, you get a $C^\infty$ function)

with \emph{domain} $\bR$, and \emph{codomain} $\bR$,
convolution is commutative; with codomain $\bC$ we make it
conjugate commutative, right?

if we're working over a non-commutative Lie group, convolution is
seriously non-commutative, obviously -- this is the Lie bracket!

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transforms coming from kernels underscore the importance of the
-improper- integral:
to take the fourier transform of $1/x$, we need to be able to make
sense of $\int_{-\infty}^\infty (\sin x)/x$ (as a limit).

if we also want to include periodic functions ``properly'' (urk...
no pun intended) and fourier coefficients in the world of fourier
transforms, then
we want the fourier transform of $\cos x$ to vanish away from $1$,
and for this we need to be able to do stuff like say
$\int_{-\infty}^\infty \sin x = 0$ in some sense.

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understanding the Fourier transform:
if you take the fourier transform of a periodic function with period
$1$, say, then you'll get (the result will be) a point-measure on
the integers, where the point-mass at each point will be the
(discrete!) fourier coefficients.
I.e., the Fourier transform of a periodic function gives a -measure-
back, not a function.

Right: letting $\hat f$ be the fourier transform of $f$,
$\hat f(x)$ (perhaps we should write $\hat f(\omega)$?)
is the amount of $f$ which has \emph{frequency} $\omega$:
note that this is the \emph{reciprocal} of wave-length.
I.e., the domain of $\hat f$ is really the frequency domain, as it's
so often called.

In particular, $\hat f(0)$ is the (improper) integral of $f$;
if $f$ has a well-defined non-negative \emph{average value} $A$
(i.e., $\lim \frac{1}{N}\int_{-N}^N f$), then $\hat f$ will have a
point-mass at the identity (corresponding to a frequency $0$ i.e.,
constant, spectral component).
(and once you subtract this point-mass $A\delta$, you'll get a value
at that point, namely the integral of $f-A$)

What are we to think of fourier transforms that only make sense for
some values? or for $\sin$ but not $\cos$?
EG, $f(x)=1/x$...this has a reasonable one for $\sin$,
and for $\cos$ it should vanish by symmetry...weird!

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there's also talk of the Fourier transform being
``rotation by $-i$ (i.e., $-90\degrees$)'' in the time-frequency domain.
(i.e., it rotates the frequency axis to the time axis...I really can't
 visualize the time-frequency domain)
(okay, I kinda lied: just write out the equation for a fractional
Fourier transform and you see what the value is at any old point:
it's $f(r,\theta)=[F_\theta f](r)$ (the value at the point
$(r,\theta)$), so something like $(f*\exp(e^{2\pi i \theta}x))(r)$)
(I still can't visualize this...)
So I guess convolving with $\exp(\frac{1+i}{\sqrt{2}}x)$ would be
rotation by $45\degrees$? -- there is a notion of fractional Fourier
transforms.
...and this explains why the inverse Fourier transform is what it is:
 it's rotation by $i$ (i.e., $90\degrees$)
...hmm...so is the $2\pi$ factor that arises really -intrinsic-?
that is, is it -monodromy- from going around a loop in the
time-frequency domain?
I.e., it's a bump that you really can't flatten.

perhaps this suggests how to understand those ``transfer-type
mappings'', as monodromy over $\bZ/2$? (considered as a category
with two objects, and just an isomorphism between the two objects:
but don't think of these as the same arrow going two different ways,
but two -different- arrows forming a loop)
(holy shit this is abstract nonsense, but it's so compelling ;-)

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??? so do we also get a "Taylor transform"?
namely something like
$Tf(\alpha)=\int f(x)x^{-\alpha}$??
(well, should take averages or something...)
(this is digusting and doesn't make much sense, if any)
(oh...the problem with this is that polynomials go off to infinity,
which is why you take derivatives rather than integrals...weird)

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hmmmm....
every measure can be -integrated-;
  if it's a finite measure, then the integral will be bounded
  variation
  (so derivatives of non-bounded variation functions should be
   seen as weird "measures"? This does matter: look at
   $\int_0^x \sin 1/t$: this function does make sense for all $x$,
     (even using proper Lebesgue integrals)
   but the derivative doesn't have bounded variation, so it doesn't
   give you a measure, right?)
but the -derivative- of a measure is not a measure:
$\delta$ is the point mass at the identity (qua measure),
and is the functional $f \mapsto f(0)$ qua distribution;
then $\delta'$ is the functional $f \mapsto f'(0)$, which seriously
doesn't make sense for general integrable functions, e.g., $\abs{x}$.
Oh....but consider the limit
$$\frac{1}{2}(\delta_\epsilon - \delta_{-\epsilon})$$
Then this converges to $\delta'$ as a \emph{distribution}
(i.e., for all $f\in C^\infty$ (indeed, $C^1$) this converges)
(I guess we'd say that it converges in the weak topology?)
but not as a measure
(i.e., not for all $f \in L^1$ (or even in $C^0$)).
Right: this is the usual topological vector space business
of restricting the space often expands the dual.

so the right way of saying it is:
measures are dual to integrable functions
  (be a bit careful: finite measures are dual to
   $L^1$ (i.e., "absolutely integrable") functions)
hence they're also dual to $C^\infty$ functions,
  i.e., they're distributions.
so we have maps:
$C^0 \to \cM \to \cD$
  (urk...actually want $L^1$ or $L^\infty$ or something mapping to
  $\cM$ -- note that the kernel is exactly functions vanishing a.e.,
  i.e., what we quotient out by to get $L^p$; I think the second map
  is injective by a Littlewood's principle: can approximate $L^1$
  by $C^\infty$)
every continuous function $f$ (on a compact set?) gives a measure
via $f \mapsto f\dx$
...okay: distributions are (by fiat) a large enough class to be closed
under taking derivatives, but they're hard to understand and have
other failings, most notably that you can't multiply them (but you can
convolve them!)
...I guess you can multiply measures (in the obvious way:
$(\mu\cdot \nu)(A)=\mu(A)\cdot \nu(A)$), and they're pretty concrete;
even more concrete are continuous functions etc.

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generality/larger classes

this (functions -> measures -> distributions) seems like the analyst
version of what we do in topology
 (expand our category from smooth manifolds to topological manifolds
  to homology manifolds, or alternatively manifolds to manifolds with
  boundary to stratified spaces),
 or algebraic geometry: varieties -> schemes -> stacks etc.
i.e., the small thing (class, category, vector space) is easily
understood and clearly of interest, while the larger collection is
less immediately accessible (and often doesn't have as special
properties), but is more tractible due to some nice properties.
Finding the "right" category is really useful
(e.g., topological spaces vs. CW-cx-able spaces vs. compactly
 generated spaces vs. spectra etc.)
...it's not just finding the right -big- category, but also about
finding interesting subcategories...

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Poisson bracket...

Lie bracket is, in these terms, $[\xi, \eta]$ where the product
is convolution!!

geometrically,
the commutativity of mixed partials
= the vanishing of the Poisson bracket
  $[\partial/\partial x_i, \partial/\partial x_j]$
= flatness of Euclidean space
(topologically it's a bit more magical and indirect:
 it's the fact that the de~Rham complex is a (co)chain
 complex -- this follows from Stokes' theorem, i.e.,
 that the derivative is the adjoint of boundary,
 but I don't see a direct topological interpretation:
 maybe I should think of it as part of the Poincar\'e lemma?)

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can differentiate w/r/t to a -vector field-,
but $\partial/\partial x_i$ doesn't make sense independently
(change of coordinates preserving one coordinate doesn't preserve
 the corresponding vector field; see it as transpose of action
 on cotangents, i.e., 1-forms)
