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A distribution is an element of the continuous dual space of some function space. Let us take the Schwartz space $\mathcal{S} := \mathcal{S}(\mathbb{R}^n)$ just as an example. A distribution $\phi \in \mathcal{S}'$ is then a map $$ \phi: \mathcal{S} \rightarrow \mathbb{C}.$$

My question is this: how do I interpret $\phi(x)$? I see this written a lot, but I don't understand how to work with it. What does for example $\phi(x) = \phi(-x)$ mean? The only thing I can think of is that $\phi(f) = \phi(\hat{f})$, where $\hat{f}(x) = f(-x)$.

And more specifically for the problem I'm working on: I have a distribution $$ \mathcal{W} : \mathcal{S}(\underbrace{\mathbb{R}^4 \times \dots \times \mathbb{R}^4}_{n \text{ times}}) \rightarrow \mathbb{C} $$

and then they say that $\mathcal{W}$ is translation invariant, i.e. $$\mathcal{W}(x_1 +a,\dots, x_n + a) = \mathcal{W}(x_1,\dots,x_n)$$ so it can be writthen as a distribution $\mathfrak{W}$ that only depends on the differences $x_1-x_2,\dots,x_{n-1} - x_n$: $$\mathcal{W}(x_1,\dots,x_n) = \mathfrak{W}(x_1-x_2,\dots,x_{n-1}-x_n).$$

How do I interpret this last line?

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When you have a $C^{\infty}$ map $f : \mathbb{R}^n \to \mathbb{R}^m$ which is proper (it is automaticaly the case with a $C^{\infty}$ bijection like translations), you can define the pushforward of a distribution by $f$. If $u$ is a distribution on $\mathbb{R}^n$ then, $f_{!}u$ is a distribution on $\mathbb{R}^m$ defined by $f_!u(\varphi)=u(\varphi \circ f).$

So for example the (dangerous) notation $u(x) = u(-x)$ means that $f_!u = u$ with $f : \mathbb{R}^n \to \mathbb{R}^n, x \mapsto -x.$

The same occurs when considering your translation by $a$. Let us call it $\mu_a.$ The condition $$\mathcal{W}(x_1 +a,\dots, x_n + a) = \mathcal{W}(x_1,\dots,x_n)$$ means that $\mathcal{W}=\mu_{a!}\mathcal{W}$

Now define the map $g : (\mathbb{R}^4)^n \to (\mathbb{R}^4)^{n-1}, (x_1, ... ,x_n) \mapsto (x_1-x_2, ... , x_{n-1}-x_n).$ The last equality means that the condition $\mathcal{W}=\mu_{a!}\mathcal{W}$ implies that $\mathcal{W}=g^*\mathfrak{W}$ for a certain distribution $\mathfrak{W}$ on $(\mathbb{R}^4)^{n-1}$.

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  • $\begingroup$ Oh, I was just trying to answer my own question with what you told me! I have that, with your map $g$, $\mathfrak{W} = g_! \mathcal{W}$, is that correct? I don't know what you mean by $g^* \mathfrak{W}$. $\endgroup$ – user353840 Jun 1 '18 at 10:08
  • $\begingroup$ You can always define the pullback of a distribution by a submersion $g$. (By duality, with integration over the fibers of $g$) Here $g$ is a submersion so the pullback has a meaning. If you want, if will edit my post in details. $\endgroup$ – C. Dubussy Jun 1 '18 at 10:54
  • $\begingroup$ I have no idea what submersion and fibers are, so if you could post details, it would be much appreciated, but I don't want to take too much of your time. :) $\endgroup$ – user353840 Jun 1 '18 at 10:56
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The trouble is that given a map $f : X \rightarrow Y$, the notation $\phi(f(x))$ should really mean "$\phi$ pulled back across $f$." But of course, distributions don't pull back in a natural way. Therefore I think the notation $\phi(f(x))$ only makes sense if $f$ is an invertible function, in which case it really means the $(f^{-1})_*(\phi),$ that is, the pushforward of $\phi$ across the inverse of $f$.

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  • $\begingroup$ Is that push forward defined as in the other answer? $\endgroup$ – user353840 Jun 1 '18 at 10:29
  • $\begingroup$ @user353840, probably. This question defines it. $\endgroup$ – goblin Jun 1 '18 at 10:36
  • $\begingroup$ @user353840, ah yes, we're defining pushforwards the same. But I disagree with that answer. $\phi(x-1)$, for example, should move $\phi$ to the right one unit. To do that, we take its pushforward across $x \mapsto x+1$. I'm pretty sure this contradicts what C. Dubbusy wrote. $\endgroup$ – goblin Jun 1 '18 at 10:39
  • $\begingroup$ How would you then do the second part? Where we use the function $g$ as in C. Dubussy's answer? Because that's not a bijection, so it doesn't have an inverse. $\endgroup$ – user353840 Jun 1 '18 at 10:49
  • $\begingroup$ Goblin is right, I should have considered the translation $\mu_{-a}$. But notice that the condition $\mathcal{W}=\mu_{a!}\mathcal{W}$ is the same as $\mathcal{W}=\mu_{-a!}\mathcal{W}$ so actually what I wrote is also correct. $\endgroup$ – C. Dubussy Jun 1 '18 at 10:59
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If the motivation is Wightman functions, it's better to use the differences $x_1-x_n,\ldots,x_{n-1}-x_n$ with a single point of reference $x_n$ rather than the consecutive differences $x_1-x_2, \ldots,x_{n-1}-x_n$. So let me consider the statement that as elements in $S'(\mathbb{R}^{4n})$ $$ \mathcal{W}(x_1,\ldots,x_n)=V(x_1-x_n,\ldots,x_{n-1}-x_n) $$ for some $V$ in $S'(\mathbb{R}^{4(n-1)})$.

First do some nonrigorous calculations as follows. Let $f(x_1,\ldots,x_n)$ be a test function in $S'(\mathbb{R}^{4n})$.

$$ \int_{\mathbb{R}^{4n}} V(x_1-x_n,\ldots,x_{n-1}-x_n)f(x_1,\ldots,x_n) \ dx_1\cdots dx_n $$ $$ =\int_{\mathbb{R}^{4n}} V(z_1,\ldots,z_{n-1})f(z_1+x_n,\ldots,z_{n-1}+x_n,x_n) \ dz_1\cdots dz_{n-1}dx_n $$ by the change of variables $z_i=x_i-x_n$ for $1\le i\le n-1$. By Fubini the last integral becomes $$ \int_{\mathbb{R}^{4(n-1)}} V(z_1,\ldots,z_{n-1})g(z_1,\ldots,z_{n-1}) \ dz_1\cdots dz_{n-1} $$ where $g=\Gamma (f)$ is defined by $$ g(z_1,\ldots,z_{n-1})=\int_{\mathbb{R}^4} f(z_1+x_n,\ldots,z_{n-1}+x_n,x_n) \ dx_n\ . $$ Now you can make rigorous sense of your RHS as the composition $V\circ \Gamma$ of the distribution $V:S(\mathbb{R}^{4(n-1)})\longrightarrow\mathbb{C}$ with the continuous linear map $$ \Gamma:S(\mathbb{R}^{4n})\longrightarrow S(\mathbb{R}^{4(n-1)})\ . $$

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