Timeline for Finite codimensional subvector space of $C^{*}$ algebras which contains no invertible elements
Current License: CC BY-SA 3.0
9 events
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| Mar 5, 2015 at 13:58 | comment | added | Ali Taghavi | Your last statment about finite dim represenation is interesting. Moreover can your argument be modified to prove $n(M_{k}(A))=kn(A)? More generally, is it true to say $n(A\otimes B)=n(A)n(B)$. Thanks again for your very interesting answer. | |
| Mar 5, 2015 at 13:55 | comment | added | Ali Taghavi | As I understand, you use a compactness argument. For every $x$, there is a neighborhood $U_{x}$ and an element $a_{x}\in A_{0}$ such that $\pi_{y}(a_{x}$ is an invertible matrices, for all $y\in U_{x}$. Compactness gives us a finite number of $a_{x_{i}}$ then you consider $\sum_{i=1}^{n} a_{x_{i}}a_{x_{i}}^{*}$, yes? | |
| Mar 5, 2015 at 13:50 | comment | added | Ali Taghavi | +1 for very elegance argument in your answer. | |
| Mar 5, 2015 at 13:47 | vote | accept | Ali Taghavi | ||
| Mar 4, 2015 at 12:40 | comment | added | Hannes Thiel | 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. | |
| Mar 4, 2015 at 12:38 | comment | added | Hannes Thiel | 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 $Y$.) | |
| Mar 4, 2015 at 8:44 | comment | added | Ali Taghavi | Forgive me for this second message. could you please more explain on your answer? | |
| Mar 3, 2015 at 18:23 | comment | added | Ali Taghavi | 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 $n-1$? | |
| Mar 2, 2015 at 10:56 | history | answered | Hannes Thiel | CC BY-SA 3.0 |