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Let $\mathcal{M}$ be a finite dimensional von Neumann algebra, then : $$\mathcal{M} \simeq \bigoplus_i M_{n_i}(\mathbb{C})$$

Question : Is it singly generated (as von Neumann algebra)? how ?

Allowed operations : $() \mapsto I$ , $(A) \mapsto \lambda A$ or $\mathbf{A^*}$ and $(A,B) \mapsto A+B$ or $A B$

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  • $\begingroup$ Given your list of allowed operations, I would replace "as von Neumann algebra" with "as a ∗-algebra over ℂ". $\endgroup$ – André Henriques Nov 23 '13 at 20:24
  • $\begingroup$ For esthetic reasons, I would replace the family of 1-ary operations $(A)\mapsto A+\lambda I$ with the 0-ary operation $()\mapsto I$. $\endgroup$ – André Henriques Nov 23 '13 at 21:21
  • $\begingroup$ @AndréHenriques : that's it ! $\endgroup$ – Sebastien Palcoux Nov 23 '13 at 21:27
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Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that for generic $\lambda_i$ that element generates your algebra.

First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by $$ A:=\left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$ and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).

Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$. The first operation is to consider $N:=A^2-aA$, which is nilpotent with non-zero terms on the subdiagonal. Then you take an appropriate linear combination of powers of $N$ to generate the element $$ \left(\begin{matrix} 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ \end{matrix}\,\right). $$ Then you take the transpose of that element, and you're essentially done.

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  • $\begingroup$ Application (after Dave Penneys): let $\mathcal{P}$ be a finite index, depth $k$ subfactor planar algebra. Then $\mathcal{P}_k$ generates $\mathcal{P}$, but $\mathcal{P}_k$ is a finite dimensional von Neumann algebra, so through your answer, $\mathcal{P}$ is generated by a single $k$-box. But look at this paper (suggested by Dave), they prove that it is generated by a single $(k+1)$-box. So your argument is better, isn't it ? $\endgroup$ – Sebastien Palcoux Nov 23 '13 at 21:36
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    $\begingroup$ Why is it interesting to know that something is singly generated? Often (e.g. in group theory), having few generators means that the defining relations become horrible, and it is better to have more generators and to keep the set of relations under control. $\endgroup$ – André Henriques Nov 23 '13 at 21:40
  • $\begingroup$ You're right, but my deep reason is that I'm looking for a translation into the planar algebra framework of the property of being cyclic for a subfactor (i.e. distributive lattice of intermediate). Note that the cyclic "group subfactors" are exactly the "cyclic group" subfactors. Bisch and Jones wrote papers on singly generated planar algebras of small dimension. Singly generated subfactor planar algebras are not cyclic in general, so I look for some restriction on the depth of the generator. $\endgroup$ – Sebastien Palcoux Nov 23 '13 at 21:51
  • $\begingroup$ Why do you take $N:=A^2-aA$ instead of $A-aI$ ? $\endgroup$ – Sebastien Palcoux Nov 23 '13 at 22:08
  • $\begingroup$ I might have overlooked something, but why can't you just take $\lambda_1=\ldots=\lambda_k=0$? $\endgroup$ – Julien Nov 23 '13 at 22:40
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For the sake of completeness, here are more general statements:

1) Every separable type I factor is singly generated. You take the weighted shift $S=\sum_j 2^{-j}E_{j,j+1}$. Then $W^*(S)$ contains $(S^*S)^{1/2}=\sum_j2^{-j}E_{jj}$ and now by using the characteristic function of $\{2^{-j}\}$ we obtain $E_{jj}\in W^*(S)$ for all $j$. Now $E_{k,k+1}=2^kE_{kk}S\in W^*(S)$ and we can get the rest of the matrix units as in $E_{k,k+2}=E_{k,k+1}E_{k+1,k+2}$, etc. Then $W^*(S)$ contains all matrix units and is then equal to $B(H)$.

2) A countable direct sum of singly generated is singly generated. If $\mathcal M=\bigoplus_{j\in J}\mathcal M_j$, $J\subset\mathbb N$, we take generators $M_j\in\mathcal M_j$ with $\sigma(M_j)\subset[3/4,1]$ (note that $M$ is a generator if and only if $\alpha M+\beta I$ is a generator for any $\alpha,\beta\in\mathbb C$). Now $M=\bigoplus_j2^{-j}M_j$ is a generator, as we can isolate each summand via functional calculus as in 1).

So 1) and 2) address Sébastien's question.

To go further, a generalization of 1) shows that any separable infinite factor (i.e. II$_\infty$ or III) is singly generated: for type II$_\infty$, you take a dense countable subset of the positive unit ball, normalize all the elements, and form the "weighted shift" as in 1). For type III, you use that $M=M\otimes B(H)$ and you can also form the weighted shift.

3) Separable abelian von Neumann algebras are also singly generated (by a selfadjoint), by a Theorem of von Neumann himself. A nice proof can be found in II.2.8 of Davidson's "C$^*$-algebras by Example".

4) Tensor product of separable abelian and singly generated is singly generated. This is a wonderful old trick. If $\mathcal A$ is separable abelian, it has a selfadjoint generator $a$. If $\mathcal M$ is singly generated, it has a generator $b+ic$ with $b,c$ selfadjoint. Then $W^*(a)\otimes W^*(b)$ is separable and abelian, and so it has a selfadjoint generator $x$; similarly $W^*(a)\otimes W^*(c)$ has a selfadjoint generator $y$. Then $x+iy$ is a generator for $\mathcal A\otimes\mathcal M$ (since $W^*(x+iy)$ contains $a\otimes I$, and $I\otimes(b+ic)$).

5) A separable von Neumann algebra with no type II$_1$ summand is singly generated. Such an algebra is of the form $\bigoplus_j\mathcal A_j\otimes\mathcal M_j$, with each $\mathcal A_j$ abelian and separable, and each $\mathcal M_j$ a non II$_1$-factor (and so, singly generated).

So the "generator problem" for von Neumann algebras is reduced to the case of II$_1$-factors.

6) The question of whether all separable II$_1$-factors are singly generated is still open.

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Take the element a=(1_{n_1},...,1_{n_k}), 1_n_i the identity matrix in M(C^{n_i}). You certainly then may generate all elements ;-).Formally: Wanting X=(X_1,...,X_k) (sum of matrices X_i, each from M(C^{n_i}), take surprisingly (X_1,...,X_k).a and you get the X. The multiplication in direct sum of algebras must be defined to make it a formal sense.

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    $\begingroup$ This is false. Notice that the element is supposed to generate the algebra as an algebra, not as an ideal. $\endgroup$ – Rasmus Nov 23 '13 at 21:04

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