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in my functional analysis class right now we are studying the basics of C* Algebras and I was recently asked this question about the second isomorphism theorem for C* Algebras, but first let me cite the definition of a C* Algebra ismorphism:

Abstract characterization of C* Algebras

Now we are asked this:

Let A be a C* Algebra and $ B \subset A $ a C* subalgebra of A and I an ideal. We are also reminded of this result: We know that B+I is a C* subalgebra of A. We are asked the following:

We are to show the following C* isomorphism: $ B/(B \cap I) \cong (B+I)/I $

I am stuck in part b as I do not really know how to show a bijective bounded (continuous) linear map which preserves multiplication and the * operation. I really need someone out there to show me how to find and prove a C*-isomorphism as desired. Also here is a related post here: sum of C*-subalgebra and ideal

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Define $\phi:(B+I)/I\to B/B\cap I$ by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If $b_1+j_1=b_2+j_2$, then $$ b_1-b_2=j_2-j_1\in B\cap I, $$ so $b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative, $*$-preserving, and onto.

As for injectivity, if $b_1+B\cap I=b_2+B\cap I$, then $j=b_1-b_2\in B\cap I$; so $$b_1+0+I=b_2+j+I.$$

Finally, $\phi$ is bounded because every $*$-homomorphism between C$^*$-algebras is. In fact, because it is a $*$-isomorphism between C$^*$-algebras, it is isometric.

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  • $\begingroup$ Thank you this does help but what about this map being continuous (or bounded as it is linear) that is where my problems are mostly? $\endgroup$ – Don John Prep Apr 29 '16 at 14:45
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    $\begingroup$ That's for free. Being a $*$-homomorphism, $\phi$ is positive. Positive maps are bounded. But more than that (and not even requiring the last sentence), bijective $*$-homomorphisms are isometric. $\endgroup$ – Martin Argerami Apr 29 '16 at 15:26

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