It's a question from HW:
Suppose we have $ \Phi:\mathbb{R}^p \to \mathbb{R}^\infty $ that satisfies:
$$ \Phi\left(x\right)^{T}\Phi\left(y\right)=\exp\left(-\frac{\left\Vert x-y\right\Vert ^{2}}{2\sigma^{2}}\right) $$ Find $ \Phi $.
First I noticed that $ \Phi(x)^T\Phi(x) = \exp(0) = 1 $. Than I substituted $ x = 0 $ and I got:
$ \Phi\left(0\right)^{T}\Phi\left(y\right)=\exp\left(-\frac{\left\Vert y\right\Vert ^{2}}{2\sigma^{2}}\right) $
Hence:
$ \Phi\left(y\right)=\Phi\left(0\right)\Phi\left(0\right)^{T}\Phi\left(y\right)=\Phi\left(0\right)\exp\left(-\frac{\left\Vert y\right\Vert ^{2}}{2\sigma^{2}}\right) $
It seems right to me but when I checked my self I found:
$ \Phi\left(x\right)^{T}\Phi\left(y\right) = \exp\left(-\frac{\left\Vert x\right\Vert ^{2}}{2\sigma^{2}}\right)\exp\left(-\frac{\left\Vert y\right\Vert ^{2}}{2\sigma^{2}}\right) = \exp\left(-\frac{\left\Vert x-y\right\Vert ^{2}}{2\sigma^{2}}-\frac{2x^{T}y}{2\sigma^{2}}\right) = \exp\left(-\frac{2x^{T}y}{2\sigma^{2}}\right)\Phi\left(x\right)^{T}\Phi\left(y\right) $.
which is not the original condition.
Not an answer, just an extended comment.
Call $q(x) = \exp(-\frac{1}{2\sigma^{2}}||x||^2)$. In the case of $p=1$, $\exp(\langle x,y \rangle / \sigma^2) = \sum_n \frac{(xy)^n}{n!\sigma^{2n}}$. Define the sequence $\Phi(x) = \left\{ \frac{q(x)x^n}{\sigma^n\sqrt{n!}} \right\}_{n=0}^{\infty}$. Then $\langle \Phi(x), \Phi(y) \rangle = \exp\left(-\frac{\left\Vert x-y\right\Vert ^{2}}{2\sigma^{2}}\right)$. It only works nicely like this I think for $p=1$.
Basically, you would need the $x$ and $y$ components to separate in $\langle x,y \rangle^n$ terms of the taylor expansion of $\exp(\langle x,y \rangle / \sigma^2)$. This doesn't really work for $p\gt 1$, which means the vector space of sequences of $p$-vectors is only the right space to look in the easy case of $p=1$.