2
$\begingroup$

My question is propably too vague to answer, the reader is advised.

There are these two mentioned theorems which you may recognise in other forms, but these are the ones that I have been taught.

Jacobson density theorem
Let $U$ be an irreducible $R$-module ($R$ is a ring with identity) and $D=End_R(U)$. Let $X\subseteq U$ be finite and $D$-free, and $\phi\in End_D(U)$. Then there is a $r\in R$ such that for all $a\in X$, we have that $\phi(a)=ra$.

Riesz representation theorem
Let $(H,\langle·,·\rangle)$ be a Hilbert space whose inner product is linear in its first argument and antilinear in the second argument. If $\phi\in H^*$, then there exists $r\in H$ such that for all $x\in H$, we have that $\phi(x)=\langle x,r\rangle$.

Is there any relation between these two theorems, apart from being two representation theorems?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ I personally don't see too much similarity. Jacobson is of the type "given a nice $A$-module $M$, the canonical $A \to End_{End_A (M)} (M)$ is surjective". Riesz is of the type "Given $V \to V^*$ with some nice properties (in particular, being injective), $V \to V^*$ is surjective". But maybe I miss something. $\endgroup$
    – Sasha
    Commented Jun 8, 2021 at 6:05
  • 1
    $\begingroup$ Is there a parallel in their corollaries? JDT (in the version you've stated above) implies that the canonical map $R \rightarrow \mathrm{End}_D(U)$ is surjective (which is how the result is often stated); Riesz implies the existence of an adjoint to an arbitrary bounded linear transformation on $H$. $\endgroup$
    – Mark Wildon
    Commented Jun 8, 2021 at 12:22
  • $\begingroup$ @MarkWildon It doesn't seem like there is a parallel. $\endgroup$
    – IJM98
    Commented Jun 8, 2021 at 12:39
  • $\begingroup$ @MarkWildon Well as the first comment from Sasha, Riesz implies also that $V\longrightarrow V^*$ is surjective, so maybe there is something. $\endgroup$
    – IJM98
    Commented Jun 9, 2021 at 0:35
  • 1
    $\begingroup$ In light of actual mathematical facts, this question turns out to be too broad, yes, but in a different universe it the answer would have been that the two things are indeed the same. So, why not here? I myself find this a fair question. :) $\endgroup$
    – paul garrett
    Commented Jun 9, 2021 at 1:24

0

Reset to default

You must log in to answer this question.