I have a small question about unitization of (unital) $C^*$-algebras. I first asked on math.stackexchange because it is basic theory, but I still have no suitable answer, the link https://math.stackexchange.com/questions/1336297/unitization-of-a-unital-c-algebra .

I will describe my problem: Let $A$ be a $C^*$-algebra, which is non-unital. Than you can prove that $$\|(a,\lambda)\|=\|L_{(a,\lambda)}\|,$$ with $L_{(a,\lambda)}:A\to A,\; b\mapsto ab+\lambda b$, defines a norm on the vector space $A_1=A\oplus\mathbb{C}$. But you need, that $A$ is non-unital to prove: if $\|(a,\lambda)\|=\|L_{(a,\lambda)}\|=0, \Rightarrow (a,\lambda)=(0,0)$.

In literature frequently it is assumed that $A$ is non-unital and then you unitize it, for example here http://ncatlab.org/nlab/show/unitization+of+a+C-star-algebra. But there is not much about the unital case in literature.

The process seems to fail, if $A$ is unital and you want to take the norm above on $A_1$. In literature, I read that you can take $\|(a,\lambda)\|=max\{\|a\|,|\lambda|\}$ in this case. But is seems curious that you have to take a different norm in this case.

My question is, is it correct to take an other norm if $A$ is unital ? But it's strange that the unitization-process depends on if $A$ is unital or not. Regards

Edit. I try to precise my question: is there a uniform construction of the norm on A_1 and a uniform proof that it really is a C*-norm which does not use the case distinction unital/non-unital?

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    $\begingroup$ Please include the link to your math.stackexchange question -- and be aware that before cross-posting you should wait a reasonable time for an answer on math.stackexchange (say, a week or two). $\endgroup$
    – Stefan Kohl
    Jun 24, 2015 at 9:04
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    $\begingroup$ @UwF: If I understand the question correctly, then "treating the case..." is exactly what causes confusion here. Why should there be any case distinction in the definitions or the proofs between non-unital algebras that just happen to have forgotten that they really have an identity and non-unital algebras that genuinely do not have one? If an abstract existence proof of the unitalization functor (e.g. from the adjoint functor theorem) works without this completely unnatural distinction, why can't there be an explicit construction of the unitalisation and its norm that does? $\endgroup$ Jun 24, 2015 at 16:06
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    $\begingroup$ Yes, of course, $A_1$ should always be bigger than $A$. Exactly because the unital/non-unital distinction on the input is completely unnatural when one wants to define a functor $\{C^\ast-algebras\}\to\{C^\ast-algebras with one\}$. Also the relation $(C_0(X))_1 = C(X^+)$ demands that, because the one-point-compactification adds a point whether or not $X$ is compact. I'll try to have look in Murphys book as soon as possible. $\endgroup$ Jun 24, 2015 at 16:27
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    $\begingroup$ I thought of taking for $A\oplus\mathbb{C}$ the $C^*$-algebraic direct sum, i.e. $\|(b,\mu)\|=\max(\|b\|,|\mu|)$, and letting $A_1$ act on $A\oplus\mathbb{C}$ as $(a,\lambda)(b,\mu)=((a+\lambda)b,\lambda\mu)$. $\endgroup$
    – UwF
    Jun 25, 2015 at 15:37
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    $\begingroup$ @DuchampGérardH.E.: Suppose some $(a,\lambda)\in A_1$ acts trivially. Then you apply it first to $(0,1)$ to see that $\lambda=0$ and then to $(a^*,0)$ to get $a=0$. I am not saying that such a construction is better than the classical proof, for this reason I don't want to turn it into an answer, either. $\endgroup$
    – UwF
    Jun 26, 2015 at 6:55

4 Answers 4


This answer is the solution UwF provided in the comments.

Let $A_1$ act on the normed space $V:=A\oplus\mathbb{C}$ with norm $\|(b,z)\|_V:=\max\{\|b\|_A,|z|\}$ via the algebra homomorphism $A_1\to B(V), (a,\lambda)\mapsto \tilde{L}_{(a,\lambda)}$ where $\tilde{L}_{(a,\lambda)}(b,z) := (ab+\lambda b,\lambda z)$. One can verify that this homomorphism is injective so that $\|(a,\lambda)\|_{A_1} := \|\tilde{L}_{(a,\lambda)}\|_{B(V)}$ is a norm on $A_1$. An easy modification of the usual proof where $A_1$ acts only on $A$ shows that this norm is indeed the $C^\ast$-Norm on $A_1$.

Nowhere in this proof is a case-distinction between unital and non-unital $A$ needed.


The question is already answer but there is a point I want to add:

Some time ago I wrote a paper about the Gelfand duality for non-unital algebra within constructive mathematics, my proof goes through the unitarization process, but in the framework of constructive/intuitionist mathematics the case distinction between unital/non-unital c* algebra is really impossible, so I had to write a proof which is completely without this distinction. It is essentially the argument outlined by Johannes Hahn (which can be found in details in my paper arxiv 1412.2009).

But it is in fact possible to construct two different unitarization process: the one described by Johannes Hahn and in my paper, and another one that uses the action of $A + \mathbb{C}$ on $A$ as in the original question.

Both are legitimate and produces $C^*$-algebras. Classically the diference between the two process is often overlooked as they agree on non-unital $C^*$-algebra and on unital algebras one gives the algebra you started with and the other gives the algebra $A^{+} = A + \mathbb{C}$. But constructively, when it is not always possible to decide wether a given algebra is unital or not, the two process are a lot more different.

This being said there is one aspect that can be seen classically on which the two constructions are very different (and to some extent it is related to the fact they are constructively different): it is their functoriality:

  • the process that change a unital algebra into $A^+$ is functorial on all algebra and all morphisms.

  • The process that preserve unital algebras is not functorial on all morphisms but only on a restricted class of morphisms, for example all non-degenerate morphisms.


Happy to find the present posts to get introduced to a unified unitisation procedure!

I'd like to elaborate further$-$thus permitting myself a higher verbosity level$-$on Johannes Hahn's answer, based on UwF's proposal (in a comment to the OP).
Let as before $V=A\oplus\mathbb{C}$ normed by $\max\{\|\cdot\|_A,|\cdot|\}$, then the $C^*$-norm on $A_1$ reads $$ \sup\big\{\max\{\|ab+\lambda b\|_A,|\lambda z|\}\;\big|\; \max\{\|b\|_A, |z|\} = 1 \big\} $$ $b$ and $z$, satisfying $\|b\|_A=1$ and $|z|=1$, may be varied independently to obtain $$ =\;\sup\big\{\max\{\|ab+\lambda b\|_A,|\lambda|\}\;\big|\; \|b\|_A = 1\big\}\; = $$

$$ \max\Big\{|\lambda|\:,\:\sup\big\{\|ab+\lambda b\|_A\,\big|\,\|b\|_A = 1\big\}\Big\}\; = \;\|(a,\lambda)\|_{A_1} $$ This expression covers both the cases "$A$ is unital" and "$A$ has no unit". The second argument to '$\max$' equals the operator norm of $L_a + \lambda\,\text{id}_A\in\mathscr{L}(A)$, where $L_a$ denotes left multiplication by $a$.

An important step consists in verifying $\forall x = (a,\lambda)\in A_1$ the $C^*$-condition $\|x\|^2_{A_1} = \|x^*x\|_{A_1}$ which is straightforward as declared in Johannes Hahn's answer: For any $b$ with $\|b\|_A=1$ one has $$ \|xb\|^2_A \le \|x^*xb\|_A \le \|L_{x^*x}\|_{\mathscr{L}(A)} $$ hence the inequality $$ \|x\|^2_{A_1} = \max\big\{|\lambda|^2,\|L_x\|^2_{\mathscr{L}(A)}\big\} \le \max\big\{|\lambda|^2,\|L_{x^*x}\|_{\mathscr{L}(A)}\big\} = \|x^*x\|_{A_1} $$ Because the operator norm is submultiplicative, one deduces $\|x\|_{A_1}=\|x^*\|_{A_1}$ which in turn proves the $C^*$-condition.

Remark: Adopting the unitisation product as the action of $A_1$ on $V$, instead of the proposed one, yields failure in the case $A =\mathbb{C}$ already:
Consider the element $(-2,1)\in A_1$. Its square is $(0,1)$, whence $$ \|(-2,1)^*(-2,1)\|_{A_1} = 1 $$ but $$ \|(-2,1)\|_{A_1}\; =\; \sup\big\{\|(-2,1)(a,\lambda)\| \big| \max\{|a|,|\lambda|\}=1\big\} $$ $$ = \;\sup\big\{\|(-a-2\lambda,\lambda)\| \big| \max\{|a|,|\lambda|\}=1\big\} \; =\; 3 $$

$\quad\Longrightarrow\; C^*$-condition is not fulfilled.

A kind of Archive
The following case distinction stems from the primary version of this answer; as it may still be helpful it is kept here:

$\quad A$ has no unit
Then the second value entering the '$\max$' in the gray block above equals the desired $C^*$-norm $$\|(a,\lambda)\|_{A_1} = \sup\big\{\|ab+\lambda b\|_A\,\big|\,\|b\|_A = 1\big\}$$

see e.g. [Pedersen: "Analysis now", 4.3.9 Lemma], also for proving the $C^*$-condition.
Since $A_1\twoheadrightarrow\mathbb{C}, (a,\lambda)\mapsto\lambda$ is a *-homomorphism, thus norm-nonincreasing, one has $ |\lambda| \le \|(a,\lambda)\|_{A_1}$. Hence considering the maximum is redundant.

$\quad A$ is unital
with unit $e$, then its unitisation $A_1$ is (isomorphic to) a direct sum of $C^*$-algebras $$A_1\overset{\cong}{\longrightarrow} A\oplus\mathbb{C},\; a+\lambda 1\longmapsto (a+\lambda e,\lambda(1-e)),$$

cf. [Murphy: "C*-Algebras and Operator Theory", 2.1.6 Thm], and $$\|(a,\lambda)\|_{A_1} = \max\big\{ \|a+\lambda e\|_A , |\lambda|\big\}$$ does the job as $C^*$-norm on $A_1$ (this rectifies a statement in the OP).

It coincides with the general expression above because (left) regular representations of $C^*$-algebras are isometric, i.e., $\|L_a\|_{\mathscr{L}(A)} = \|a\|_A$ holds $\forall a$ in any $A$
(to show equality: "$\le$" only depends on the norm being submultiplicative,
whereas "$\ge$" results from choosing $a^*/\|a\|$ if $a\ne 0$, combined with the $C^*$-condition).
Making the operator norm definition explicit within the current situation yields $$\|a+\lambda e\|_A = \sup\big\{\|(a+\lambda e)b\|_A\,\big|\,\|b\|_A = 1\big\}$$


If fact, if you apply the unitization process on an already unital $C^*$-algebra (say $\mathcal{A}$), then the "old" unit $1_\mathcal{A}$ remains an idempotent (this is precisely the projector on $\mathcal{A}$ parallel to $\mathbb{C}1_{new}$) but ceases to be the unit (this rôle being taken by the fresh unit). $$ \mathcal{A}_1=\mathcal{A}\oplus \mathbb{C}1_{new} $$ and in the sector $\mathcal{A}$, the new norm coincides with the old one.

Now it is clear that your question can be made rigorous with the help of categories. Let us call $\mathbf{C^*}$-alg (resp. $\mathbf{C^*}$-alg$_\mathbf{1}$) be the category of $C^*$-algebras with or without unit (resp. the category of $C^*$-algebras with unit) with morphisms all $*$-homomorphisms between them (resp. all unital $*$-homomorphisms between them), it seems to me clear that the unitization solves the following universal problem

Given $\mathcal{A}$ in $\mathbf{C^*}$-alg, provide $\mathcal{A}_1$ and an arrow $\alpha : \mathcal{A}\rightarrow F(\mathcal{A}_1)$ ($F$ being the forgetful functor $\mathbf{C^*}$-alg$_\mathbf{1}$ towards $\mathbf{C^*}$-alg) such that for all arrow $\beta : \mathcal{A}\rightarrow F(\mathcal{B})$ ($\mathcal{B}$ is unital), there exists a unique morphism $\beta_1 : \mathcal{A}_1\rightarrow \mathcal{B}$ of $\mathbf{C^*}$-alg$_\mathbf{1}$ which fulfills $\beta=F(\beta_1)\alpha$.

As usual the solution (if it exists) is unique up to isomorphism. To show the existence we construct one particular solution and this needs two cases. This comes from the fact, (general in algebra) that, in the case when $A$ has already a unit, the representation of $A_1$ within the operator algebra of $A$ by left multiplication is no longer faithful. So, as to define the norm, we need a faithful representation, we only consider it in the case without unit and compute it directly in the case when $A$ is already unital. This is well illustrated in Bourbaki Spectral Theory Chapter 1 Paragraph 6 Proposition 2 unitization, the statement says "there is a unique norm ..." (in all cases, this answers your edit but the construction is not uniform due to the problem of faithfulness). Hope it helps.

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    $\begingroup$ This doesn't answer any of the questions, doesn't it? $\endgroup$ Jun 24, 2015 at 16:32
  • $\begingroup$ Does anybody care to explain this downvote ? $\endgroup$ Jun 25, 2015 at 2:45
  • $\begingroup$ I don't see how this addresses the issue of existence (or construction) of the left adjoint between the relevant categories $\endgroup$
    – Yemon Choi
    Jun 25, 2015 at 12:49
  • $\begingroup$ This is stated in the (grey) block, the problem is stated as "for each object of $\mathcal{C}$, provide an object of $\mathcal{C}_1$ and an arrow ... " do not hesitate to ask clarification ... $\endgroup$ Jun 25, 2015 at 13:33
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    $\begingroup$ @JohannesHahn There is a big misunderstanding here. I always wrote that the Bourbaki statement is uniform and the proof is case studying. The action of UwF is a good idea as I said, but I am not shure that if you write the proof completely (evaluation of the norms) you do not have to return to the case study (well, I have done it although) [You said : Nowhere in this proof is a case-distinction between unital and non-unital] (a) If it is the case it is very good news (b) If it is your choice could you explain your downvote ? $\endgroup$ Jun 26, 2015 at 16:06

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