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I have a little question about unitization of a $C^*$-algebra. If $A$ is a non-unital $C^*$-algebra, set $A_1=A\oplus\mathbb{C}$ as vector spaces and define a multiplikation, involution and a norm in a usual way, $$(a,\lambda)\cdot(b,\mu)=(ab+\mu a+\lambda b,\lambda\mu)$$ $$(a,\lambda)^*=(a^*,\overline{\lambda})$$ and $$\|(a,\lambda)\|=\|L_{(a,\lambda)}\|,$$ with $L_{(a,\lambda)}:A\to A,\; b\mapsto ab+\lambda b$ is the multiplication operator. You will obtain a $C^*$-algebra $A_1$ with unit $(0,1)$ and $A$ can be identified with $\{(a,0);a\in A\}$, $A$ is a closed ideal in $A_1$. My question is, what exanctly changes if $A$ is unital and you consider it's unitization $A_1$?

If $A$ is unital, its unitization $A_1$ is again $A\oplus\mathbb{C} $ as vector spaces and the involution is the same. But the norm seems to be different, and why?
What else is different in this case?

Edit: Ok, my problem is tho following:

Frequently, if the unitization of $A$ will be introduced, it is assumed that A is non-unital, for example here: http://ncatlab.org/nlab/show/unitization+of+a+C-star-algebra . But if I want to show that $\|(a,\lambda)\|=\|L_{(a,\lambda)}\|$ is a norm, I need that A is non-unital to prove: if $\|(a,\lambda)\|=\|L_{(a,\lambda)}\|=0, \Rightarrow (a,\lambda)=(0,0)$.

Therefore my question is, what is the correct $C^*$-morm on $A_1$, if $A$ is unital? For example I read $\|(a,\lambda)\|=max\{\|a\|,|\lambda|\}$ should be the $C^*$-Norm on $A_1$ if A is unital. But I dont see how it could be, because $\|(a,\lambda)\|=max\{\|a\|,|\lambda|\}$ is completely different as $\|(a,\lambda)\|=\|L_{(a,\lambda)}\|$.

To put my question in a nutshell: What exactly is the $C^*$-norm on $A_1$ if $A$ is unital and non-unital?

An other thing is the multiplication in the non-unital case and the unital case. If $A$ is a Banach algebra, the multiplication on $A_1$ should be $(a,\lambda)\cdot(b,\mu)=(ab+\mu a+\lambda b,\lambda\mu)$, but I read that if $A$ is unital, than the multiplication is $(a,\lambda)\cdot (b,\mu)=(ab,\lambda\mu)$. But I don't know why..

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The issue is not about the norm. If $A$ is already unital, then the unitization construction as you described it is not well-defined, in the following sense. Since $1\in A$, you can consider the element $(-1,1)$, which of course should be zero. This forces you to construct the unitization as a quotient $(A\oplus\mathbb C)/J$, where $J$ is the ideal $$ J=\{(-t,t):\ t\in \mathbb C\}. $$ And then you can easily show that $(A\oplus\mathbb C)/J\simeq A$.

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  • $\begingroup$ ok, this meens, in every case, whether A is unital or not, the $C^*$-norm and the multiplication defined on $A_1$ is the same? I will edit my question $\endgroup$ – google-hupf Jun 23 '15 at 18:43
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    $\begingroup$ Why should (-1,1) be zero? It shouldn't! Otherwise you wouldn't get the relation $(C_0(X))_1 = C(X^+)$. The one-point compactification of an already compact space still adds one point, so the C*-algebra should get strictly larger too. $\endgroup$ – Johannes Hahn Jun 24 '15 at 15:57
  • $\begingroup$ But then the construction of the unitization is not the one mentioned in the original question. And I'm not sure how canonical it is to have a compactification where the original space is not dense. $\endgroup$ – Martin Argerami Jun 24 '15 at 21:18
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Same issue arises in plain Banach algebras (except then you can be a little sloppier defining the norm on the unitization). If $A$ is a Banach algebra with identity and you define $A_1$ as above then $(0,1)$ is the identity in $A_1$. The previous identity $e$, which looks like $(e,0)$ in $A_1$, is no longer an identity, because for example $(e,0)(0,1)=(e,0)$.

You ask why the norm seems to be different. Not sure what you mean - if you mean the norm on $A_1$ is different from the norm on $A$, yes it is. It's a different space. If you mean the norm works out differently in this case from how it works out for non-unital $A$ I don't see what difference you're referring to.

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  • $\begingroup$ ok, thanks. sorry I was not precise enough. But your answer is helpful too and I will vote up. I will edit my question $\endgroup$ – google-hupf Jun 23 '15 at 18:37

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