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](https://mathoverflow.net/questions/215501/an-equivalent-relation-on-the-space-of-polynomials-in-one-complex-variable)


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](https://en.wikipedia.org/wiki/Quotient_space_(topology)#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?
> 
> 
> 



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


