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I'm fairly new to $C^*$-algebras and Hilbert space. Given the algebraic relations of the $C^*$-algebra, I am having a lot of trouble computing the norm of its elements and am wondering if there are tricks I am missing. For example: suppose $a$ is a partial isometry: $aa^*a=a$ (and $a^*aa^*=a^*$). And suppose there are no other relations. Let $c$ be a complex number. What is $\|ca+\bar{c}a^*\|$? I've tried two ways to figure this out. One (since this is self-adjoint) is to compute the spectrum directly and hence find the spectral radius. I end up with recursion relations for the coefficients of terms; these are quotients of polynomials. Then I find the roots of the denominators. I was able to guess from this that the answer is 2|c|, but this was just a guess from finding a sequence of denominator polynomials whose roots seemed (numerically) to converge to that. This is also a ton of work for such a simple example. So then I tried making a model in sequence space. Here I think is an example of such an $a$: it sends vector $v=\{v_0,v_1,v_2\ldots\}$ to $$\{v_0,0,v_3,0,v_6,0,v_9,0,v_{12},0,\ldots\}$$ and then $a^*$ sends $v$ to$$\{v_0,0,0,v_2,0,0,v_4,0,0,v_6,0,0,\ldots\}$$It is easy then to write down $ca+\bar{c}a^*$. But what is the norm of this operator? I couldn't figure it out by staring at it. The best I could do was to truncate the infinite operator matrix to an $n\times n$ matrix and compute the eigenvalues for particular values of $c$, then let $n$ get large. Which seemed to very slowly converge towards my guess answer, but who knows? Any help appreciated.

Edit: Martin helped me see that I made a mistake when I described my example by saying that $a$ was a partial isometry "with no other relations." What I intended was a lot more restrictive than that: I wanted to require that all products of $a$ and $a^*\!$ (that is: $a$, $a^*\!$, $aa$, $aa^*\!$, $a^*\!a$, $a^*\!a^*\!$, and so forth) are partial isometries. It follows that these terms form the free inverse semigroup generated by $a$. (I wonder if there is a simpler way to describe this situation than to say that all those products are partial isometries? Such as: a finite number of conditions on $a$?) If $a$ is the operator on sequence space given earlier, all of the above terms are equal to one another iff the rules of inverse semigroups say they should be equal, and terms that are not equal are linearly independent. (In retrospect it seems lucky that I found an operator that behaves as I wanted it to, given that there are so many other ways for a partial isometry to be.)

I examined the characteristic polynomials of the truncated matrices of my example operator, and thereby realized that the infinite matrix is the direct sum of (many copies of) matrices that look like this:$$\left(\begin{array}{cccc}0&c&0&0\\\bar{c}&0&c&0\\0&\bar{c}&0&c\\0&0&\bar{c}&0\end{array}\right)$$that is: $c$'s on one off-diagonal, $\bar{c}$'s on the other, and zeros everywhere else. But of arbitrary size. So the problem of finding the norm of my operator reduces to the problem of finding the sup of the norms of these matrices. Numerically it seems that the result is $2|c|$, but I don't know how to prove that.

David Hillman
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Fix $\lambda\in\mathbb C$ with $|\lambda|=1$ and $\lambda c=|c|$. Then $$ a_0=\begin{bmatrix} \lambda &0\\0&0\end{bmatrix} $$ is a partial isometry, and $$ \|ca_0+\overline ca_0^*\|=|c\lambda+\overline c\overline\lambda|=2|c|. $$ Now, if you want to assume that $a$ lives in a C$^*$-algebra where it has no other relations, then $C^*(a)$ is the universal C$^*$-algebra generated by a partial isometry. So given any partial isometry $a_0$, there exists a $^*$-epimorphism $\pi:C^*(a)\to C^*(a_0)$ with $\pi(a)=a_0$. As $*$-homomorphisms are contractive, $$ \|ca+\overline c a^*\|\geq\|c\pi(a)+\overline c\pi(a^*)\|=\|ca_0+\overline c a_0^*\|=2|c|. $$ As we also have $\|ca+\overline ca^*\|\leq2\|ca\|=2|c|\,\|a\|=2|c|$, we get $$ \|ca+\overline c a^*\|=2|c|. $$ Note that $\|a\|=\|a^*a\|=1$, since $a^*a$ is a projection.

The example you attempted is not universal. For instance $a^2(a^*)^2a^2=a^2$.

How do we know that the universal algebra exists?

Let $A$ be the free $*$-algebra on one generator $a$. For any C$^*$-algebra $C^*(b)$ generated by a partial isometry $b$, there exists a $*$-representation $\pi:A\to C^*(b)$ given by $\pi(p(a,a^*))=p(b,b^*)$ for any non-commutative polynomial $p$. This is well-defined since all monomials in $A$ are linearly independent. For $x\in A$, let $$ \|x\|=\sup\{\|\pi(x)\|:\ \pi:A\to C^*(b)\ \text{ is a $*$-homomorphism and $b$ is a partial isometry }\}. $$ This seminorm is well-defined (that is, it is finite for each $x$) since for any polynomial $p$, the norm of $p(b,b^*)$ has a bound independent of $b$, due to $\|b\|=1$.

Let $K=\{x\in A:\ |x\|=0\}$. This is an ideal, and so $\|x+K\|=\|x\|$ defines a norm on $A/K$. Now put $C^*(a)=\overline{A/K}$, the completion. This is the universal C$^*$-algebra generated by a partial isometry, since for any partial isometry $b$ we have a $*$-epimorphism $\pi_b:C^*(a)\to C^*(b)$ with $\pi(a)=b$.

Indeed, by the first paragraph in this section a $*$-homomorphism $\pi:A\to C^*(b)$ always exists. And by definition of the norm on $A/K$, $$ \|\pi_b(x)\|\leq\|x\|, $$ which makes $\pi_b$ a contraction. This allows us to extend $\pi_b$ to its completion $C^*(a)$.

Martin Argerami
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  • I think you missed my sentence "And suppose there are no other relations." Here your first example satisfies $a^2=0$ and your second example satisfies $\lambda^2 a^=a$. I'm looking for a partial isometry such that the terms made by multiplying $a$'s and $a^$'s together correspond to linearly independent operators unless required to be equal by the rules of partial isometry and the rules of $C^*$-algebras. Isn't it true that in that case its norm would be uniquely determined? – David Hillman Oct 08 '18 at 22:20
  • That's not a very good definition, it seems, since it's hard to predict which terms should be linearly independent and which shouldn't. The canonical way to talk about those things is to require $C^*(a)$ to be the universal algebra generated by a partial isometry: I've edited my answer to reflect that. – Martin Argerami Oct 10 '18 at 05:34
  • Thank you, that is so helpful! (Contractive! Told you I was a beginner.) I will try to understand your universal construction. Meanwhile, on the topic of $aaa^a^aa$: I may have been making a false assumption (which I blame on being used to inverse semigroups). In a $C^$-algebra is it not automatically true that for a partial isometry $a$ we have $aa^=a^a$? (My intent was that products of $a$ and $a^$ follow the axioms of an inverse semigroup, so perhaps I needed to say more than that it is a partial isometry.) – David Hillman Oct 11 '18 at 15:40
  • No, you shouldn't expect partial isometries to be normal. For the easiest example, take $a=\begin{bmatrix}0&1\0&0\end{bmatrix}\in M_2(\mathbb C)$. – Martin Argerami Oct 11 '18 at 16:04
  • I'm sorry, I misspoke: I meant $aa^$ commutes with $a^a$, not $aa^=a^a$. But I guess that I cannot expect that either? – David Hillman Oct 11 '18 at 17:02
  • No, you can't. For instance take $a=\begin{bmatrix} 1/\sqrt2&0\1/\sqrt2&0\end{bmatrix}$. – Martin Argerami Oct 11 '18 at 17:09
  • To give you some intuition, if your C$^$-algebra is $M_n(\mathbb C)$ or $B(H)$, given any two projections $p,q$ with equal rank you can find $a$ with $a^a=p$, $aa^*=q$. This $a$ is simply the map that takes an orthonormal basis of the range of $p$ to an orthonormal basis of the range of $q$. – Martin Argerami Oct 11 '18 at 17:26
  • I've worked on understanding your section "How do we know that the universal algebra exists?" (I think there is a typo: shouldn't the definition of $K$ say $|x|=0$ rather than $\pi(x)=0$?) I think I see why the construction gives a $C^!$-algebra, but don't yet see how to show the epimorphism part. And am confused by thoughts like: since the norm in $C^(a)$ is bigger than the norm in $C^*(b)$, couldn't a series converge in the latter that doesn't converge in the former? – David Hillman Oct 16 '18 at 02:12
  • I added a last paragraph. I don't know why you talk about series; what matters is that $\pi_b$ is contractive and thus can be extended. – Martin Argerami Oct 16 '18 at 03:00