# $H^{*}$ algebras as a generalization of $C^{*}$ algebras

Let $$H$$ be the quaternions algebra. An $$H^{*}$$ algebra is a normed ring $$A$$ which is simultaneously a unital left $$H$$ module and has an involution $$*$$ with the following properties:

$$\forall \lambda \in H, a,b \in A$$

1.$$\;\lambda(ab)=(\lambda a)b$$

1. $$\; \parallel ab\parallel \leq \parallel a \parallel \parallel b \parallel,\;\;\; \parallel \lambda a\parallel=\parallel \lambda \parallel \parallel a\parallel$$

2. $$\;(ab)^{*}=b^{*}a^{*}$$

4.$$\;\;\parallel ab\parallel \leq \parallel a \parallel \parallel b \parallel$$

1. $$\;\; \parallel aa^{*} \parallel= \parallel a \parallel^{2}$$

2. (If A is unital) $$\lambda 1 \lambda' 1= \lambda \lambda' 1$$

There is a natural definition of spectrum of an element $$a\in A$$ (as a subset of $$H$$). There is also a natural definition of morphism and isomorphisms between two $$H^{*}$$ algebras.

Example: For a compact Hausdorff space $$X$$ put $$A=H(X)=$$ The space of all continuous $$f:X \to H$$ with obvious structures

Questions:

1. Is it true to say that the spectrum is always non empty and compact?

2. Is it true to say that two compact space $$X$$ and $$Y$$ are homeomorphic if and only if $$H(X) \simeq H(Y)$$?

The motivation for this question is that we search for an alternative proof for the Borsuk Ulam theorem for $$f;S^{4} \to \mathbb{R}^{4} \simeq H$$ via consideration of a $$\mathbb{Z}/2\mathbb{Z}$$ graded structure for $$H(S^{4})$$. see the following related post

Banach algebraic proof of the Borsuk Ulam theorem

Edit: According to the answer of Andre Henriques we ask:

1. Assume that $$A$$ is a real $$C^{*}$$ algebra which contains the Quaternions $$H$$. Under what algebraic conditions, $$A$$ is in the form $$H(X)$$=The space of continuous functions $$f:X \to H$$ for some compact Hausdorff $$X$$?
• Warning: the terminology $H^*$-algebra has been used in the past for something completely different, although I guess it is not so commonly studied nowadays – Yemon Choi Oct 20 '15 at 20:14
• @eric $H$ is not a complex algebra – Ali Taghavi Oct 20 '15 at 20:40
• I think that your notion of an $H^*$-algebra is the same thing as a real $C^*$-algebra equipped with a copy of the quaternions in it. – André Henriques Oct 20 '15 at 21:36
• @AndréHenriques May I suggest you upgrade your comment to an answer? – Theo Johnson-Freyd Oct 20 '15 at 23:35
• @AliTaghavi : I'm surprise you don't ask to have an action by $H$ on the right as well. Of course you could define the action on the right by $x \lambda = (\lambda^* x^* )^*$ but then I think you need an axiom to obtain that $\lambda (x \lambda') = (\lambda x) \lambda'$. – Simon Henry Oct 21 '15 at 8:31

I think that your notion of an $H^*$-algebra is essentially the same thing as a real $C^*$-algebra equipped with a copy of the quaternions in it.

But you should probably add the axiom which says that $1+a^*a$ is invertible in (the unitalization of) $A$. That axiom is not needed for complex $C^*$-algebra, but it is needed for real $C^*$-algebras.

• This only work in the unital case, and even for that there is some axioms missing in the Op definition for this to work. In the general case you could say with a copy of the quaternion within the Multiplier algebra but this for example already contains the action of $H$ on the right which is missing from the OP definition. The condition $(1+a^*a)$ is part of the definition of real $C^*$-algebra and it can be replace by the strong $C^*$-inequality ($\Vert x^* x + y^* y \Vert \geqslant \Vert x \Vert ^2$). – Simon Henry Oct 21 '15 at 10:01
• It's still unclear to me how you get property (2), that $\| ax\| = \| a\| \|x\|$ for $a\in\mathbb H$, if you just know that $\mathbb H$ sits inside your $C^*$ algebra. Could you please elaborate. – Christian Remling Oct 21 '15 at 15:58
• @ChristianRemling : It is also true for real $C^*$-algebra that a injective morphism of real $C^*$-algebras is isometric, hence if $H$ sit inside $A$ you have that $\Vert h \Vert$ is indeed what it is suppose to be, the fact $\Vert hx \Vert = \Vert h \Vert \Vert x \Vert$ follow from the fact that $h$ has an inverse whose norm is $1/\Vert h \Vert$ and two application of the multiplicativity of the norm. Of course one need $H$ to sit inside $A$ as a sub $*$-algebra, which include an additional axiom on the behavior of $*$ on $H$ in comparison with what the OP is asking. – Simon Henry Oct 21 '15 at 16:06
• @Andre thank you for your answer. What about the spectrum of elements(non empty compact subsets of H?) – Ali Taghavi Oct 22 '15 at 6:59
• Indeed, I followed a suggestion by Theo Johnson-Freyd to upgrade my comment to an answer. – André Henriques Oct 27 '15 at 16:20

Let assume that you consider unital algebra only (one can still study non unital algebra by unitarizing them, but notion of spectrum is always a little annoying when one want to consider non unital algebra) and that a $H^*$ algebra is a real $C^*$-algebra with morphism of real $C^*$-algebra $\mathbb{H} \rightarrow A$

First two remarks:

• The name $H^*$ is IMHO not a good idea at all, first the $C$ in $C^*$-algebra is not for "complex", and second $H^*$-algebra are actually already a thing, see for example here. I would rather call them $\mathbb{H}$-$C^*$-algebra or maybe quaternionic $C^*$-algebras (as we say real $C^*$-algebras and not $R^*$-algebras)

• This definition might not be completely satisfying in the sense that it does not generalizes complexe $C^*$-algebras: a complexe $C^*$-algebra is the same as a real $C^*$-algebra together with an injection of $\mathbb{C}$ into its center. but when you replace $\mathbb{C}$ by $\mathbb{H}$ you need to remove the assumption on the center (because $\mathbb{H}$ is non-commutative) and it is not clear that it is a good idea to just remove it. For example, it might be a good idea to add all sort of commutativity conditions that holds in $\mathbb{H}$, like " if $x$ is self adjoint then $x$ commute to all elements of $\mathbb{H}$ which would simplify some answer below...

The only part I don't know how to answer is the non-emptyness of the spectrum... I'm still thinking about it, and it might be related to those possible additional condition I'm mentioning just above. I will edit if I find something.

So, let $A$ be a quaternionic $C^*$-algebra, and $a \in A$ I guess you want to define the spectrum as $\{ h \in \mathbb{H} | a - h \}$ is non-invertible.

Then the proof that is compact is exactly the same as in the usual case: you prove that it is bounded and closed using that if $X$ is invertible and $Y$ is such that $\Vert Y \Vert < 1/\Vert X^{-1} \Vert$ then $X+Y$ is invertible using a series argument (that work for arbitrary complete normed algebra)

As I said, 'non-empty' is trickier because the spectrum in a real $C^*$-algebra don't have to be inhabited and we don't have holomorphic calculus at disposition for quaternionic $C^*$-algebra (althoug it might be an idea to develop it a little, the resolvant is still locally a formal series so maybe one can use a kind of quaternionic Liouville's theorem). But it seems difficult to produce a counter example: For example, as soon as you have an element which commute to some purely imagnary unit $u$, then $x,x^*,u$ generates a complexe $C^*$-algebra (with $u$ as $i$) hence $x$ has a non-empty spectrum inside of it, and it implies that $x$ has a non-empty sepctrum (as in real $C^*$-algebra it also true that $x$ is invertible is a subalgebra if and only if it is invertible in the larger algebra: it is proved by taking the complex form of the algebras).

For your question 2, the usual proof caries over easily and one can see that $X$ is the space of $\mathbb{H}$-algebra morphism from $H(X)$ to $\mathbb{H}$ hence if $H( X) \simeq H(Y)$ then $X \simeq Y$ and the isomorphism between $H(X)$ and $H(Y)$ is induced by the homeomorphism between $X$ and $Y$..

For your question 3), you always have a morphism from $A$ to $H(Spec A)$ where $Spec A$ is the set of character of $A$ (i.e. morphisms of $\mathbb{H}$-algebra from $A$ to $\mathbb{H}$). So "the algebras for which this morphism is an isomorphism" is an answer to your question. of course one can hope to obtain something better... If I'm correct the quaternionic Stone-Weierstrass theorem work, and the only unital $\mathbb{H}$-*-sub-algebra of $H(X)$ which separates points is $H(X)$ itself, hence the comparison map $A \rightarrow H(Spec A)$ is always surjective.

But one can give a more explicit condition: A quaternionic $C^*$-algebra is of the form $H(X)$ if and only if its self adjoint element ($x^*=x$) are central (or just commute between themselves and to elements of $\mathbb{H}$).

Indeed, it is then obvious that the set of self-adjoint element is a real $C^*$-algebra with trivial involution hence that it is of the form $C(X,\mathbb{R})$ for some $X$.

Moreover, using a kind of polarization formula, one can write any element in a unique way in the form $x+iy+jz+kt$ where $x,y,z,t$ are self-adjoint, and because self-adjoint are central, those multiply exactly as function on $X$ with values in $H$.

• @SimonHenrey I really thank you very much for your answer. Is it obvious that Spec A is non empty? – Ali Taghavi Oct 27 '15 at 14:22
• I don't think it is in general (it is not the case for ordinary $C^*$-algebra either), under the "commutativity" condition (that self-adjoint are central) then yes it is because it is the spectrum of a non zero real $C^*$-algebra with trivial involution. – Simon Henry Oct 27 '15 at 14:35
• I don't know a good reference of real C* algebra: they are a bit scattered through the literature (trivial involution just mean that there $*$-operation is the identity), I found this papper arxiv.org/pdf/1505.04091v1.pdf on google that seem to give a bit of introduction and lots of references and might be a good place to start (and its recent so its bibliographie is probably up to date). What you need to know is that when $C$ is a real $C^*$-algebra then $C \otimes \mathbb{C}$ is an ordinary $C^*$-algebra endowed with an involution coming from complex conjugation and that... – Simon Henry Oct 27 '15 at 14:58
• Then it is also true that the only division algebras over $\mathbb{R}$ are $\mathbb{R}, \mathbb{C}$ and $\mathbb{H}$, hence the only division quaternionic $C^*$ algebra is $\mathbb{H}$. – Simon Henry Oct 28 '15 at 10:02
• I wasn't talking about Wedderburn, but the real form of Gelfand Mazur theorem that assert that the only real banach division algebra are $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$ (sorry, I forgot to say banach in my previous comment) I don't know any standard reference for this either, but google give me this ams.org/journals/proc/1995-123-09/S0002-9939-1995-1231298-6/… it contains a proof and references... – Simon Henry Oct 29 '15 at 10:08