Connection of branches

Question

1. I can't understand the meaning of kernel function in a Fourier space. What is the relation between the Fourier kernel $K(w,t)$ and homomorphism?

2. Please explain the connection of Mathemetics with: the Fourier, group, Linear-Algebra and Integral calculus.

Answer 1

The connection between branches of mathematics you are seeking is best given as a response to your second question. Alas, I will provide a response to both questions to clear up, what seems to be, confusion due to terminology.

question 1). First, lets refresh ourselves of the defintion of a homomorphism. A homomorphism is a map between two algebraic structures of the same type, that preserves the operations of the structures. This means a map

$f$ : $A \to B$

between the sets $A$ and $B$ equipped with the same structure such that, if $\bullet$ is an operation of the structure (which we will assume, for simplification, to be a binary operation), then

$f(a_{1} \bullet a_{2}) = f(a_{1}) \bullet f(a_{2})$, $\forall a \in A$.

With that in mind, the kernel of a homormpism is the subet $Ker_{f} :=$ {$a \in A$: $f(a) = e_{B}$} $\subseteq A$ where $e_{B}$ is the identity element in $B$. I can see how this idea of the kernel as a set can be confused with the kernel of an integral since this terminology overlaps with the fact that, for example, the definite integral is is a homomorphism. Observe: let $P$ be a set whose elements are polynomials, therefore there exists a map $I : P \to \mathbb{R}$ such that $I(p)$ $:= $$\int_0^1$$p(x)dx$. However, some authors call the argument $p(x)$ of the integral $I$ the kernel of the integral. Therefore with our definition of the map $I$, there exists a subset $Ker_{I} \subseteq P$ such that

$Ker_{I}:=$ {$p \in P$: $\int_0^1$$p(x)dx$ $= 0$}.

The overlap in terminology here is that $Ker_{I}$ is the kernel of the homomorphism $I$, but the argument of the integral (that defines $I$) according to some literature is considered "the kernel of the integral".

So in the context of the the Fourier integral transform $T$ of the following form

$(Tf)(u)$ $:= $$\int_{t_{1}}^{t_{2}}$$f(t)K(t,u)dt$

by the explanation above, there is no real connection between the Fourier kernel $K(t,u)$ and the kernel of a homomorphism besides a sloppy use of terminology.

question 2). The connection between linear algebra, group theory, and integral calculus is that in all three branches we study homomorphisms. Whether or not its between vector spaces or algebraic structures. In particular, in integral calculus we study maps like $I$ that we defined above from the set $P$ to the real numbers. One thing to obviously note is, as i'm sure you've noticed, we use integral calculus when studying Fourier transformations.