# $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m) \simeq \mathcal S(\mathbb R^{n+m})$?

If $S(\mathbb R^n)$ is the Scwartz space of smooth rapidly decaying functions equipped with the topology generated by the family of semi-norms $$\mathcal N_p (\varphi)= \sum_{|\alpha|, |\beta| \leq p} \sup_{x\in \mathbb R^n} |x^\alpha \partial^\beta \varphi (x) |\, ,$$ Is it true that $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)\simeq \mathcal S(\mathbb R^{n+m})$, where $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)$ is the completion of $\mathcal S(\mathbb R^n) \otimes \mathcal S(\mathbb R^m)$ for the topology $\pi$ defined by the family of semi-norms $$\mathcal N_{p,q} (\varphi\otimes \psi)=\mathcal N_p(\varphi)\mathcal N_q (\psi)\, .$$

• As in @PeterMichor's answer, the nuclearity means that genuine tensor products exist, as opposed to the more typical situation that one can have one half or the other, but not both, of the characterizing properties of a tensor product. Some of the notes on my functional analysis page math.umn.edu/~garrett/m/fun talk about the impossibility of having a "genuine" tensor product of Hilbert spaces, and also about nuclear Frechet spaces occuring as suitable (proj) limits of Hilbert spaces. – paul garrett Mar 27 '15 at 23:16

• Thank you very much! The topology on the tensor product is then generated by the family of semi norms $\mathcal N_p$ defined above? – Thomas Mar 30 '15 at 8:08
• Grothendieck called the $\varepsilon$ tensor topology injective. The inductive topology is still finer and even for nuclear spaces it may be different from the injective one. – Jochen Wengenroth Apr 6 '15 at 13:31