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Assume that $A$ is a unital $C^{*}$ algebra. Is there a subvector space $Y\subset A$ of finite codimension which does not contain any invertible element?

Let $n(A)$ be the infimum of such codimensions. For example $n(A)=1$ if $A$ is commutative. Or $n(M_{n}(\mathbb{C}))=n$.

For a commutative $A$, is it true to say $n(A\otimes M_{n}(\mathbb{C}))=n\times 1=n$? More generaly, can one express $n(A\otimes B)$ in term of $n(A)$ and $n(B)$?

Edit: According to the comments on this post, we ask:"What is an example of a simple $C^{*}$ algebra $A$ for which $n(A)$ is finite? Of course for such algebra, $n(A)\neq 1$.

Note: This question is somehow a reverse question to the following famous problem:

What is the maximim possible dimension for those subvector space of $M_{n}(\mathbb{R})$ which consist only invertible matrices(except zero matrix). For what valuse of $n$ the sharp dimension would occure?

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    $\begingroup$ Have you considered your question for $A = B(H)$, with $H$ an infinite-dimensional Hilbert space? $\endgroup$
    – Manny Reyes
    Commented Feb 7, 2015 at 19:08
  • $\begingroup$ @MannyReyes thank you for you comment and your revision. No, I do not know how to try this question for $B(H)$. There is no any determinant type function. $\endgroup$
    – Ali Taghavi
    Commented Feb 7, 2015 at 19:20
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    $\begingroup$ @MannyReyes according to your comment, I think the answer is negative for $B(H)$ since each $M_{n}(\mathbb{C}$ is embeded in $B(H)$. On the other hand this codimension goes to infinity as $n\to \infty$. Am I right? $\endgroup$
    – Ali Taghavi
    Commented Feb 7, 2015 at 19:39
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    $\begingroup$ this is a good point about $B(H)$, as long as you truly know that $n(M_n(\mathbb{C}) = n$. (I see that it's $\leq n$, but not quite equality.) Regarding $n(A) = 1$, note that for any (not necessarily commutative) unital C*-algebra $B$ one has $n(B \times \mathbb{C}) = 1$. $\endgroup$
    – Manny Reyes
    Commented Feb 7, 2015 at 20:37
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    $\begingroup$ One small observation is that $n(A)=1$ if $A$ has a character. How did you get $n(M_n(\mathbb C))\geq n$? $\endgroup$
    – Rasmus
    Commented Feb 10, 2015 at 8:31

1 Answer 1

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Let $A$ be a unital C*-algebra. As was already noted in the comments, we have $n(A)=1$ if and only if $A$ has a character.

Let $M_n=M_n(\mathbb{C})$. It is easy to see that $n(M_n)\leq n$. Conversely, a linear subspace of $M_n$ of codimension less than $n$ contains an invertible matrix, as shown by Dieudonné, [1], see also [2]. Thus $n(M_n)=n$.

If $\pi\colon A\to B$ is a surjective ${}^*$-homomorphism onto another C*-algebra $B$, then $n(A)\leq n(B)$. Indeed, let $B_0$ be a subspace of $B$ of codimension $n(B)$ such that $B^{-1}\cap B_0=\emptyset$. Set $A_0:=\pi^{-1}(B_0)$. Then $A_0$ has codimension $n(B)$ in $A$ and $A^{-1}\cap A_0=\emptyset$. Therefore $n(A)\leq n(B)$.

Next, let $X$ be a compact, Hausdorff space and consider $A=C(X,M_n)=C(X)\otimes M_n$. We show $n(A)=n$.

First, for every $x\in X$, there is a surjective ${}^*$-homomorphism $\pi_x\colon A\to M_n$ given by evaluating at $x$. It follows $n(A)\leq n$. Let us show that the converse inequality also holds. So assume there exists a subspace $A_0$ of noninvertible functions of codimension $\leq n-1$. Then there are at most $n-1$ points in $X$ such that $\pi_x(A_0)\neq M_n$, say $x_1,\ldots,x_k$. For each $j\in 1,\ldots,k$, since $A_0$ has codimension $\leq n-1$, we have that $\pi_{x_j}(A_0)$ has codimension $\leq n-1$, and so $\pi_{x_j}(A_0)$ contains an invertible matrix. Since the set $\{x_1,\ldots,x_k\}$ is discrete in $X$, we can easily construct an invertible element in $A_0$, a contradiction.

One could now ask if $n(A)<\infty$ if and only if $A$ has a (nonzero) finite-dimensional representation (equivalently, a nonzero, finite-dimensional quotient C*-algebra).

[1] Dieudonné: Sur une généralisation du groupe orthogonal à quatre variables, Arch. Math. 1, 282-287 (1949)

[2] Fillmore, Laurie, Radjavi: On matrix spaces with zero determinant, Linear multilinear algebra 18, 255-266 (1985)

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  • $\begingroup$ thank you very much for your answer. I would appreciate if you more explain on the last paragraph, from "Then there are at most $n-1$ points in $X$............, a contradiction". I do not understand the details. In particular "Since $A_{0}$ has codimension $\leq n-1$ we have that $\pi_{x_{j}}(A_{0})$ has codimension at most $n-1$? $\endgroup$
    – Ali Taghavi
    Commented Mar 3, 2015 at 18:23
  • $\begingroup$ Forgive me for this second message. could you please more explain on your answer? $\endgroup$
    – Ali Taghavi
    Commented Mar 4, 2015 at 8:44
  • $\begingroup$ More generally: If $\varphi\colon X\to Y$ is a surjective linear map between vector spaces, and $X_0$ is a subspace of $X$ of codimension $n$, then $Y_0:=\varphi(X_0)$ is a subspace of $Y$ of codimension at most $n$. Indeed, let $x_1,\ldots,x_n\in X$ such that every $x\in X$ can be written as $x=x_0+\sum_k c_k x_k$, for $x_0\in X_0$ and coefficients $c_k$. We claim that the analogous statement holds in $Y$, with $Y_0$ and the vectors $y_k:=\varphi(x_k)$, $k=1,\ldots,n$. (This will show that $Y_0$ has codimension at most $n$ in $Y$.) $\endgroup$
    – Hannes Thiel
    Commented Mar 4, 2015 at 12:38
  • $\begingroup$ So let $y\in Y$. Since $\varphi$ is surjective, there exists $x\in X$ such that $y=\varphi(x)$. By assumption, there exist $x_0$ and $c_k$ such that $x=x_0+\sum_k c_k x_k$. Using that $\varphi$ is linear, it follows $y=\varphi(x)=\varphi(x_0)+\sum_k c_k y_k$. Since $\varphi(x_0)\in Y_0$, the claim is proved. $\endgroup$
    – Hannes Thiel
    Commented Mar 4, 2015 at 12:40
  • $\begingroup$ +1 for very elegance argument in your answer. $\endgroup$
    – Ali Taghavi
    Commented Mar 5, 2015 at 13:50

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