I wonder if every maximal two-sided (self-adjoint) ideal in a C*-algebra is automatically closed. It is a very basic fact of C*-algebra theory that it holds true for the unital case. In the non-unital case, there certainly exists a non-closed dense ideal but the point is that such an ideal may never be maximal. It is a non-trivial fact that the answer is affirmative for the commutative case [D. Rudd, On isomorphisms between ideals in rings of continuous functions. Trans. Amer. Math. Soc. 159 (1971) 335--353].

One can ask a similar question for maximal *-subalgebras.

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    $\begingroup$ Is the answer known for separable C*-algebras? $\endgroup$
    – Nik Weaver
    May 25, 2017 at 13:59
  • $\begingroup$ @Nik Weaver: Not to my knowledge. I don't know about Banach algebras either. This problem occurred to me sometime ago when I was giving a lecture on functional analysis and C*-algebras. $\endgroup$ May 25, 2017 at 23:11
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    $\begingroup$ In the commutative case it is not hard: note first that maximal modular ideals in commutative Banach algebras are maximal and closed. So we need to deal with the non-modular case. However, a maximal ideal $J$ in a commutative Banach algebra $A$ is not modular if and only if $A\cdot A\subseteq J$ and $J$ has co-dimension 1 in $A$. As C*-algebras factor, every maximal ideal in a commutative C*-algebra is modular. $\endgroup$ May 27, 2017 at 16:11
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    $\begingroup$ Hi Tomek. You seem to be using the fact maximal ideals in a CBA always have codimension 1? I was unware of this fact until reading a note of HGD from around 2014, and he seemed to indicate that he did not find this result in the standard sources $\endgroup$
    – Yemon Choi
    May 29, 2017 at 22:11
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    $\begingroup$ For those following along, the HGD note is arxiv.org/abs/1408.3815 $\endgroup$ May 30, 2017 at 8:25

2 Answers 2


Theorem: Let $J\subseteq A$ be a maximal ideal. Then $J$ is hereditary (if $a\in A_+$ satisfies $a\leq b$ for some $b\in J_+$, then $a\in J_+$), strongly invariant (if $x^*x\in J_+$ then $xx^*\in J_+$) and if $a\in J_+$ then $a^t\in J_+$ for every $t>0$.

We need some notation and some lemmas first:

We define the relation $\sim$ on $A_+$ by setting $a\sim b$ if there exists $x\in A$ such that $a=xx^*$ and $b=x^*x$.

Lemma 1: If $a\leq b\sim b'$, then there exists $a'$ such that $a\sim a'\leq b'$.

Proof: Let $x\in A$ such that $b=x^*x$ and $b'=xx^*$. Let $x=v|x|$ be the polar decomposition in $A^{**}$. Set $y:=va^{1/2}$ and $a':=yy^*$. Then $y$ belongs to $A$ and we have $a=y^*y$ and $a'=yy^*=vav^*\leq vbv^*=b'$.

Lemma 2: Let $a,b,c\in A_+$ and $t>1$ such that $a\sim b\leq c^t$. Then there exists $y\in A$ such that $a=ycy^*$.

Proof: Choose $x\in A$ such that $a=xx^*$ and $b=x^*x$. Set $\alpha:=\tfrac{1}{2t}$. Then $0<\alpha<1/2$. Applying the polar decomposition in C*-algebras (see Proposition~II.3.2.1 in Blackadar's Operator algebras book) we obtain $y\in A$ such that $x = y(c^t)^\alpha$. Then $$ a = xx^* = yc^{2t\alpha}y^* = ycy^*. $$

Proof of Theorem: We consider the ideal as suggested by @Black $$ K := \{ a\in A : a^*a \leq b \text{ for some } b\in J_+ \}. $$ We will show that $J=K$. Indeed, as noted by @Black, we either have $J=K$ or $K=A$. We show that $K=A$ leads to a contradiction.

So assume that $K=A$. Let $a\in A_+$. Since $a^{1/4}\in K$, we obtain $b\in J_+$ such that $a^{1/2}\leq b$. Then $$ a = a^{1/4}a^{1/2}a^{1/4} \leq a^{1/4}ba^{1/4} \sim b^{1/2}a^{1/2}b^{1/2} \leq b^2. $$ By Lemma 1, we obtain $a'$ such that $a\sim a'\leq b^2$. It then follows from Lemma 2 that $a\in J_+$. Thus $J=A$, a contradiction.

With an argument as in the answer of @Black, it now follows that $J$ is hereditary, strongly invariant, and closed under roots of positive elements.

  • $\begingroup$ Wow! This is a very nice argument. I've been struggling to rule out (ii) as it does look pretty unlikely, but without success. It is great that you did it so nicely! So now, in order to answer the original question, one may add hereditary to the hypothesis! $\endgroup$
    – Black
    Oct 10, 2020 at 21:44
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    $\begingroup$ This is nice! This immediately yields that every maximal ideal $I$ in a commutative C*-algebra $A$ is closed. Indeed, since any $a\in A_+\setminus I$ becomes an order unit of $A/I$ by maximality and $-a \notin A_+ + I$ by hereditariness, the Hahn-Banach (Eidelheit–Kakutani) separation theorem provides a nonzero positive linear functional on $A$ that vanishes on $I$ and so $I$ cannot be dense. $\endgroup$ Oct 12, 2020 at 0:16
  • $\begingroup$ @Narutaka This is really fantastic! I wonder if this justifies publishing a short note somewhere, perhaps in the PAMS? $\endgroup$
    – Black
    Oct 12, 2020 at 14:35
  • $\begingroup$ @Hannes, I just made a few corrections in your statement and proof of Lemma 1, but since I do not have enough reputation, this might take some time to show. I guess you need to change $b$ to $b'$ at the end of the statement and set $b=x^*x$ and $b'=xx^*$ in the proof. $\endgroup$
    – Black
    Oct 12, 2020 at 14:46
  • $\begingroup$ Related to @Narutaka's idea, I've just noticed that the self-adjoint part of $A_++I$ coincides with $\{a\in A_{\text{sa}}: a_-\in I\}$. I wonder if this is of any use. $\endgroup$
    – Black
    Oct 20, 2020 at 17:21

Since this question has been around for over three years without an answer, I suppose it is OK to give a partial answer even if the advance is tiny.

THEOREM: Let $J$ be a maximal ideal in a C*-algebra $A$. Then either

(i) $J$ is hereditary and for every $a\in J_+$, and every $r>0$, one has that $a^r\in J$, or

(ii) $A_+$ coincides with its order-ideal generated by $J_+$.

PROOF: It is easy to see that $M:= \{m\in A:\exists b\in J,\ m^*m\leq b\}$ is an ideal containing $J$. Given that $J$ is maximal we conclude that $M=J$ or $M=A$. In the first case we claim that $J$ is hereditary. To see this suppose that $0\leq a\leq b$, where $a\in A$ and $b\in J$. Setting $m=a^{1/2}$, we then have that $$ m^*m = a \leq b \in J, $$ so $m$ is in $M$, hence also in $J$, so $a=m^2\in J$, proving that $J$ is hereditary.

Still under the assumption that $M=J$, let us prove the last part of (i). For this let $a\in J_+$ and put $m=a^{1/2}$. Then $$ m^*m=a\leq a\in J, $$ so we see that $m\in M=J$. In other words $a^{1/2}\in J$. By iterating this procedure we conclude that $a^{1/2^n}\in J$, for all $n$. Given $r>0$, pick a sufficiently large integer $n$ such that $1/2^n<r$. Then $$ a^r=a^{1/2^n}a^{r-1/2^n}\in J.$$

In the second case, namely if $M=A$, we will prove that the order-ideal generated by $J_+$ coincides with $A_+$. To see this, let $a$ be an arbitrary positive element in $A$. Then $m:= a^{1/2}$ lies in $M$ by hypothesis, so there is some $b$ such that $$ a= m^*m\leq b\in J, $$ proving that $a$ lies in the order-ideal generated by $J_+$, as desired.

REMARK. Not all ideals in a C*-algebra are hereditary. See Is every 2-sided ideal in a C*-algebra hereditary?

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    $\begingroup$ Could you clarify how this addresses the question "is J closed"? $\endgroup$
    – Yemon Choi
    Oct 1, 2020 at 16:24
  • $\begingroup$ I think that condition (ii) cannot occur. I will post this as an answer, since the argument is too long to fit here. $\endgroup$ Oct 10, 2020 at 19:59

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