# Fredholm subvector spaces of $B(\mathcal{H})$

Let $\mathcal{H}$ be a separable Hilbert space with orthonormal base $\{e_i\}\;\;i\in \mathbb{N}$.

Definition: We say a subvector space $W\subset B(\mathcal{H})$ is a Fredholm subspace if there is a constant $M$ such that $Ind(T)\leq M$ for all operators $T \in W$ which are Fredholm. In the other words $W$ is a Fredholm subspace if the Index is a bounded function on $$Fred(W):=\{T\in W \mid\; \text{T is a Fredholm operator\}}$$ According to this definition $W$ is NOT counted as a Fredholm subspace if $W$ contains no Fredholm operator, at all.

Example: Put $\mathcal{H}=\ell^{2}$. $S_1$ is the shift operator on $\ell^{2}.$

We define $W=\{P(S_1)\mid P\;\; \text{is a polynomial of degree at most n\}}$. Then $W$ is a $n+1$ dimensional Fredholm subspace of $B(\ell^2)$. See this post

In this question we would like to ask "Is the space of Fredholm subspaces an open set?"

We try to give a meaning to the latter statement via Grassmanian in $B(\mathcal{H}):$

We define an inner product on $B(\mathcal{H})$ with $<A,B>=\sum \frac{1}{n^2}<Ae_n,Be_n>$. This obviously enable us to define the Grassmanian $G(n, B(\mathcal{H})),$ the space of all $n$ dimensional subvector space of $B(\mathcal{H})$ with a natural topology as follows:

Let $S$ be the unit sphere of $B(\mathcal{H})$ with respect to the norm arising from the above inner product. So $S$ has a natural topology. We define a unique topology on $G(n, B(\mathcal{H}))$ such that the following map be a quotient map $$Span:\{(x_1,x_2,\ldots,x_n) \in \overbrace{S\times S\times \ldots\times S}^{n-times}\mid <x_i,x_j>=0\}\to G(n, B(\mathcal{H}))$$

This map sends $(x_1,x_2,\ldots,x_n)$ to the $n$ dimensional subspace $W$ generated by $x_1,x_2,\ldots,x_n.$

So our question is the following:

Is the space of all Fredhom subspaces of $G(n,B(\mathcal{H})$ an open set? What about if we replace $S$ by unit sphere of $B(\ell^{2})$ with its operator norm and its natural topology?

• No vector subspace of $B(H)$ can consist only of Fredholm operators since $0$ is not a Fredholm operator. – Liviu Nicolaescu Aug 24 '15 at 17:38
• @LiviuNicolaescu I think there is a misunderestanding. Can I ask you to read the definition again?Thank you. – Ali Taghavi Aug 24 '15 at 17:40
• I permute two words to remove the misunderstanding. – Ali Taghavi Aug 24 '15 at 17:42
• @LiviuNicolaescu Now is the definition, clear? – Ali Taghavi Aug 24 '15 at 17:45

The statement is false for $n= 2$, as I'll show, with respect to both distances you are considering on $G(n,B(H))$ (this easily implies that it is also false for any $n>2$; on the other hand for $n=1$ the statement is true according to your definition of Fredholm subspace, as it reduces to the fact that Fredholm operators are an open subset of $B(H)$. )
Let $S$ be the injective shift operator on $\ell^2$. Recall that for any $\lambda\in \mathbb{C}$ the operator $S-\lambda$ is injective and Fredholm of index $-1$ if $|\lambda|<1$, it is not Fredholm if $|\lambda|=1$, and it is invertible if $|\lambda|>1$. It will be of use a countable direct sum of operators $\lambda_j S$; to do it more nicely, we can realize it on a suitable Hilbert decomposition of $\ell^2$ into countably many infinite dimensional subspaces.
So let $\alpha:=(\alpha_k)_{k\ge1}$ be a bounded sequence of complex numbers. Consider the bounded operator $L^\alpha$ on $H:=\ell^2(\mathbb{Z}_+)$ defined on the standard orthonormal bases $(e_k)_{k\ge1}$ by $$L^\alpha e_k = \alpha_k e_{2k}\ .$$ Assume further that the sequence $\alpha$ satisfies $\alpha_{2k}=\alpha_k$ for all $k\ge 1$. Then $H$ splits in a Hilbert sum of a countable family of $L^\alpha$-invariant subspaces, $H=\bigoplus_{j\in\mathbb{N}} H_j$, with $H_j:=\operatorname {Span}(e_{(2j+1)2^n}: n\ge0)$. Moreover, for any $j\ge0$ the operator $L^\alpha_{\ |H_j}$ on $H_j$ is unitary equivalent to $\alpha_{2j+1}S$ on $\ell^2$. Thus for any $\lambda\in\mathbb{C}$ the operator $L^\alpha-\lambda I$ is unitary equivalent to the direct sum $\bigoplus_{j\in\mathbb{N}} \big(\alpha_{2j+1}S-\lambda\big)$. The $j$-th component $\alpha_{2j+1}S-\lambda$ is either non-Fredholm, or injective Fredholm of index $-1$, or invertible according whether $|\lambda|=|\alpha_{2j+1}|$, $|\lambda|<|\alpha_{2j+1}|$, respectively $|\lambda|>|\alpha_{2j+1}|$, in which case $\big\|\big(\alpha_{2j+1}S-\lambda\big)^{-1}\big\|\le\frac{1}{|\lambda|-|\alpha_{2j+1}|} .$
Therefore the operator $L^\alpha-\lambda I$ is Fredholm of index $$\mathrm{ind}(L^\alpha-\lambda I)=-\mathrm{card} \{j\ge0\ : |\alpha_{2j+1}| > |\lambda| \}\ ,$$
provided $|\alpha_{2j+1}|$ is bounded away from $|\lambda|$ and $|\alpha_{2j+1}|<|\lambda|$ for all but finitely many $j$, say $m$: indeed, in this case $L^\alpha-\lambda I$ is the direct sum of a bounded family of invertible operators whose inverses have bounded norms, plus $m$ Fredholm operators of index $-1$, thus the direct sum of an invertible operator and a Fredholm operator of index $-m$.
As a consequence, the $2$-dimensional subspace $V^\alpha$ generated by the operator $L^\alpha$ and the identity $I$ is a Fredholm subspace according to your definition, if e.g. the sequence $|\alpha_{2j+1}|$ is eventually constant, and it is not, if e.g. it is strictly decreasing. Indeed, in the former case the index of Fredholm operators in $V^\alpha$ can only assume finitely many values, while in the latter case $V^\alpha$ possesses Fredholm operators of any negative index. On the other hand, for bounded sequences $\alpha$ and $\beta$ one has $\|L^\alpha-L^\beta\|\le |\alpha-\beta|_\infty$ for the operator norm, so that $V^\alpha$ depends continuously from $\alpha$ (and a fortiori, if you adopt the inner product norm deduced from $\langle\ ,\ \rangle$, which is weaker). Since an eventually constant sequence is arbitrarily close to strictly decreasing sequences in the $\ell_\infty$ norm, we conclude that Fredholm subspaces of $G(2,B(H))$ are not an open subset.