User manoj - MathOverflow

Universal definition of Fourier transform

2010-10-15T21:13:52Z

Is there a category theoretic definition for the Fourier transform using only its universality properties? I am not looking for the most general definition -- one that works only in some special settings will do. I am looking for a simple definition that will make precise my (possibly incorrect) intuition that Fourier transforms are in some sense extremal among unitary transforms.

Here is another non category-theoretic way to ask this question, which may or may not be equivalent: Give a "natural" optimization problem on the space of unitary transforms whose solution turns out to be the Fourier transform.

Answer by Manoj for When is $\ker AB = \ker A + \ker B$?

2011-11-06T07:04:32Z

Here's another statement of more or less the same result. The ideas in the proof are from this proof in German that Martin Brandenburg linked to in his comment.

Claim: Let $n$ be a non-negative integer. Let $A$, $B$ be two $n×n$ square matrices over the complex numbers. If $AB=BA$ and $\ker A \cap \ker B = {0}$ then $\ker AB=\ker A \bigoplus \ker B$.

Note: Assuming that $\ker A\cap\ker B={0}$ is not a big restriction, since we can always quotient out to eventually reduce to this case.

Proof: Since $A,B$ commute, it is clear that $\ker A \bigoplus \ker B\subseteq \ker AB$. Further, $\ker AB$ is invariant under $A, B$. Since all the action is taking place within $\ker AB$, we may assume without loss of generality that $\ker AB$ is the entire space, of dimension $n$. This implies $\operatorname{im} B\subseteq \ker A$.

By the rank-nullity theorem, $\dim\ker B + \dim\operatorname{im} B = n$. Hence, $\dim\ker A + \dim\ker B\geq n$. Since these spaces intersect trivially by assumption, we are done.

When is $\ker AB = \ker A + \ker B$?

2011-11-05T08:48:20Z

Prove/ Disprove: Let $n$ be a positive integer. Let $A$, $B$ be two $n \times n$ square matrices over the complex numbers. If $AB = BA$ and $\ker A = \ker A^2$ and $\ker B = \ker B^2$ then $\ker AB = \ker A + \ker B$.

(Recall that $\ker A$ is the set of all vectors $v$ such that $Av = 0$.)

Background: I am teaching linear algebra this semester. I did not like the standard proof of the Jordan canonical form I found in the textbooks, and thought I could prove it differently, directly from the axioms for a vector space, without using either the determinant, or the classification theorem for finite abelian groups. If the statement above is true, I believe I have a proof for the Jordan canonical form for $T$ by setting $A = (T-\lambda_1I)^{n_1}$ and $B=(T-\lambda_2I)^{n_2}$ for appropriate $n_1$ and $n_2$.

Note 1: If $A = B = \left( \begin{array}{cc} 0 & 1\\0 & 0 \end{array} \right)$ then $AB = BA$ but $\ker AB \neq \ker A + \ker B$.

Note 2: It is easy to find $A, B$ such that $\ker A = \ker A^2$ and $\ker B = \ker B^2$ and $B$ maps a vector outside $\ker A + \ker B$ to $\ker A$, so that $\ker AB \neq \ker A + \ker B$.

Hence, both conditions are necessary.

Must finite groups with isomorphic commutators and quotients be isomorphic?

2010-05-29T17:31:09Z

Let G and H be finite groups. Let G' = [G,G] and H' = [H,H] be the corresponding derived groups (commutator subgroups) of G and H. I am looking for an example where G' is isomorphic to H' and G/G' is isomorphic to H/H' but G and H are not isomorphic.

When and why did the postdoctoral position originate?

2010-03-02T17:56:54Z

Does anyone know when and how the system of post-doctoral studies after a Ph. D. originated? I've heard in a few places that there was a time when there was no such thing as a post-doc, and people used to go for faculty positions right after their Ph. D.'s. Is this true? When and why did it change? What does the post-doctoral position achieve?

Zariski open sets are dense in analytic topology

2010-02-17T11:23:35Z

How does one show that if $U \subseteq \mathbb{C}^n$ is nonempty and Zariski open then $U$ is also dense in the analytic topology on $\mathbb{C}^n$?

Comment by Manoj

2011-11-06T03:21:26Z

Thanks, Mark, quite a nice proof.

Comment by Manoj

2011-11-06T03:13:15Z

So you seem to have proved something stronger: If $AB=BA$ and either $\ker A = \ker A^2$ or $\ker B=\ker B^2$ then $\ker AB=\ker A+\ker B$.

Comment by Manoj

2011-11-06T03:11:31Z

Thanks, Konstantin. It is a very neat proof! I notice that you did not make use of the assumption that $\ker B^2 =\ker B$.

Comment by Manoj

2011-11-06T03:10:45Z

Thanks, Andrew. Actually that's the book I'm teaching from. He does avoid determinants, but he essentially goes through the classification theorem for finite abelian groups (putting matrices into upper triangular form: Thm 8.10). I wanted to avoid that.

Comment by Manoj

2011-11-05T11:51:12Z

6. Therefore, $dim(\ker h(\gamma)) + dim(\ker g(\gamma)) >= dim W$. But hold on, these kernels intersect trivially, as we already showed. So we're done!

Comment by Manoj

2011-11-05T11:51:01Z

3. $\alpha$ commutes with $W=\ker (gh)(\alpha)$, so $W$ is invariant under $\alpha$. Let $\gamma$ be the restriction of $\alpha$ to $W$. We have already shown that the two other kernels of interest live inside $W$. 4. Restricted to $W$, we know that $(gh)(\gamma)$ is identically zero. Therefore the image of $h(\gamma)$ must be contained in the kernel of $g(\gamma)$. [The proof seems to get this backwards. Perhaps they are using left multiplication?] 5. But NOW! $dim(\ker h(\gamma)) + dim(im h(\gamma)) = dim W$ by the rank-nullity theorem! 6. Therefore, $dim(\ker h(\gamma)) + dim(\ker g(\ga

Comment by Manoj

2011-11-05T11:50:15Z

Thanks, Martin. Satz 1 would certainly give me the kind of proof I am looking for. If I'm not mistaken, it says that: Claim: If g,h are polynomials in one variable whose gcd is 1, then for every endomorphism $\alpha$, the kernel $\ker (gh)(\alpha)$ is a direct sum of $\ker g(\alpha)$ and $\ker h(\alpha)$. Proof: 1. If there is a vector $v$ in the intersection of the kernels of $g(\alpha)$ and $h(\alpha)$, then its annihilator (in the polynomial ring) is trivial, so $v=0$. 2. Proves the easy direction of the inclusion: $\ker g(\alpha) \subseteq \ker (gh) (\alpha)$. 3. $\alpha$ commutes wi

Comment by Manoj

2010-10-16T15:15:24Z

Thanks for the comments. To be honest, I am groping towards the right question. I am perfectly happy working even in finite-dimensional real vector spaces. The degree of generality is not the point (though it would be nice if the definition generalizes.) By universality properties, I mean limits/ colimits of some diagrams in this category. I am trying to get at a definition of the Fourier transform that is "global" in character, and is formulated in a coordinate-free manner.

Comment by Manoj

2010-02-19T06:51:48Z

Thank you Pete for the correction.