The n-Category Café

Being more of a discrete guy, I wonder about the Discrete Fourier Transform. The continuous statement becomes “cyclic convolution exchanges with pointwise multiplication” for the DFT. But in this case this is equivalent to the statement that the unitary DFT matrix of a given size diagonalises all circulant matrices of that size. I wonder if there are any other discrete transforms that are characterised by their actions on certain classes of matrix.

David wrote:

Is there a reason we pick one for the transform and the other for the inverse transform?

Not a strong reason; it’s just a matter of notation, or convention. But it’s still interesting. In what follows, I’ll use the variable $x$ where you used $t$, thinking of my waves as functions on space rather than functions of time, because people often use somewhat different sign conventions for Fourier transforms in the space and time variables — in part for good reasons (special relativity), and in part just for tradition.

Typically the variable $k$ is used as the Fourier transform partner of the space variable $x$, while $\omega$ is used as the partner of the time variable $t$. The name for $k$ is wavenumber (or sometimes (spatial) frequency), while $\omega$ is called (temporal) frequency. There are a bunch of further subtleties which I won’t bother mentioning now. They may seem purely terminological, and to some extent that’s true — but I claim that they’re all important, even if you’re mainly interested in the math!

A function of the form

$$f(x) = e^{i k x}$$

is said to be a plane wave of wavenumber $k$. Suppose we’d like our Fourier transform to map any function of this sort to a delta function supported at $k$. This will let us take a take a function take its Fourier transform, and easily see the frequencies of the plane waves from which its built as a superposition. To do this, we need to use this Fourier transform:

$$\widehat{f}(k) = \frac{1}{2 \pi} \int f(x) e^{-i k x} \, d x$$

It’s a bit easier to see that the corresponding inverse transform must be

$$f(x) = \int \widehat{f}(k) e^{i k x} \, d k$$

since this clearly sends a delta function supported at some value of $k$ back to the corresponding plane wave $e^{i k x}$.

The whole point of this overly long comment is this:

The minus sign and the factor of $2 \pi$ need to be put in the right places to make my above statement true: if we randomly muck with them, the Fourier transform of $e^{i k x}$ won’t be a delta function supported at $k$.

Anyone who wants to fully understand the Fourier transform needs to understand this, and also understand the various other conventions people use, and what’s good about those.

For example: the Fourier transform I’ve described is not a unitary operator. To make it unitary, we can adjust it as follows:

$$\widehat{f}(k) = \frac{1}{\sqrt{2 \pi}} \int f(x) e^{-i k x} \, d x$$

and then its inverse is

$$f(x) = \frac{1}{\sqrt{2 \pi}} \int \widehat{f}(x) e^{i k x} \, d x$$

This is a good convention for quantum mechanics, and it’s somehow pleasing that the necessary factor of $1/2 \pi$ has been split evenly between the Fourier transform and its inverse.

On the other hand, there are those (especially highbrow mathematicians) who prefer the kernel $e^{2 \pi i x}$ to the kernel $e^{i x}$, based on the philosophy that $2 \pi i$ is the fundamentally important number, while $2 \pi$ is less so and $\pi$ is downright dumb. And this too has its merits.

Indeed one could write a very nice essay on the various conventions for Fourier transforms, the endless arguments that people get into about them, and what this means about the sociology and philosophy of mathematics, physics and engineering. I feel one coming on, but I’ll stop here.

John said

On the other hand, there are those (especially highbrow mathematicians) who prefer the kernel $e^{2\pi i x}$ to the kernel $e^{i x}$, based on the philosophy that $2\pi i$ is the fundamentally important number, while $2\pi$ is less so and $\pi$ is downright dumb. And this too has its merits.

One aspect of this is how you think of (or parametrize) the circle. Do you parametrize by arc-length in which case you think of it as $\mathbb{R}/2\pi \mathbb{Z}$? Then the kernel is $e^{-i n x}$. Do you set the length of your circle to be $1$, which is like setting the period of periodic functions to be $1$, so you think of the circle as $\mathbb{R}/\mathbb{Z}$? Then the kernel is $e^{-2\pi i n x}$. Or do you think of the circle as the unit complex numbers? Then the kernel is $x^{-n}$.

On the other hand, there are those (especially highbrow mathematicians) who prefer […] based on the philosophy that $2\pi i$ is the fundamentally important number, while $2 \pi$ is less so and $\pi$ is downright dumb.

The highbrow mathematician will hopefully know that if there is anything “downright dumb” here then it is decreeing any fixed choice at all.

Add this to the list of examples in reply to Minhyong Kim’s request for why it is good to remember isomorphisms and not impose equality.

Urs wrote:

The highbrow mathematician will hopefully know that if there is anything “downright dumb” here then it is decreeing any fixed choice at all.

As a corollary to that, it is also downright dumb for authors to assume their readers know all the authors’ conventions without having them spelled out.

David wrote:

Likewise with the Laplace transform, why $e^{−k t}$ one way and $e^{k t}$ back?

This is again mainly convention, but this choice matters more, psychologically at least, than the choice between $e^{i k t}$ versus $e^{- i k t}$.

The choice between $e^{i k t}$ versus $e^{- i k t}$ is the choice between a counterclockwise spiral and a clockwise one. There are three easy ways to switch between these choices:

• replacing $i$ with $-i$ (Galois group symmetry),
• replacing $t$ with $-t$ (time reversal symmetry), or
• replacing $k$ with $-k$ (changing conventions about the definition of ‘frequency’).

The choice between $e^{k t}$ and $e^{-k t}$ is the difference between exponential growth and decay. This seems like a bigger deal. Now we only have two easy ways to switch between these choices: switching the sign of $t$ and switching the sign of $k$.

The Fourier transform is best for systems that oscillate endlessly. The Laplace transform is best for systems that show exponential growth or decay. In electrical circuits containing resistors, or the heat equation, or radioactive decay, or many other dissipative linear systems, we see a lot of exponential decay and no exponential growth. These are the natural markets for the Laplace transform.

For thermodynamic reasons that can be endlessly discussed but seem intuitively obvious, we have chosen our arrow of time to point forwards in the ‘decay’ direction. So, the only remaining freedom concerns our choice of the sign for $k$: do we describe exponential decay by $e^{-k t}$ with $k$ positive, or $e^{k t}$ with $k$ negative?

Thanks. That’s helpful. I wonder why we don’t see more of Mackey’s full Laplace transform, i.e., the one arising from his generalized characters. Do we not have a need to treat decaying oscillations?

See how I try to lure you back to temperature lives on the Riemann sphere.