# Extension of $*$-representation from an algebraic corner of a $*$-algebra

Let $$B$$ be a $$*$$-algebra and $$A\subseteq B$$ a $$*$$-subalgebra. Let $$p\in B$$ be a projection such that

$$pBp=A.$$

Suppose we have a $$*$$-homomorphism $$\phi:A\rightarrow\mathcal{B}(H)$$, where $$H$$ is some Hilbert space.

Question: Can $$\phi$$ be extended to a $$*$$-homomorphism $$\phi':B\rightarrow\mathcal{B}(H)$$?

Not necessarily. Let $$B= M_{2\times 2}(\Bbb C)$$ and let $$p=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$. Then $$A$$ are those matrices where only the $$11$$ component is non-zero. Let $$\varphi: A\to \mathcal B(\Bbb C)=\Bbb C$$ be the map sending such a matrix to its only non-zero component. This is a $$*$$-algebra morphism.
There can be no extension of this, because no $$M_{2\times 2}(\Bbb C)$$ admits no characters (meaning no $$*$$-algebra morphisms into $$\Bbb C$$). If you denote with $$\varphi_{ij}$$ the image of the matrix with a $$1$$ on the $$ij$$ component and else $$0$$ of such an extension, you have $$(\varphi_{12})^2=0=(\varphi_{21})^2$$ since this is an algebra morphism, implying $$\varphi_{12}=\varphi_{21}=0$$. But $$\begin{pmatrix}0& 1\\ 1&0\end{pmatrix}$$ is invertible, so either $$\varphi_{12}+\varphi_{21}$$ is invertible (not possible) or $$\varphi$$ is the zero morphism.
• Thanks for your answer. I wonder if the answer would be different if we only allow $H$ to be infinite-dimensional? – ougoah Jul 23 at 1:24
• Yes, take the same $A,B$ and let $H=\ell^2$ and $\varphi_{11}=\mathrm{id}$. You must have $(\varphi_{12})^2=0=(\varphi_{21})^2$ but $\varphi_{12}\varphi_{21}=\varphi_{11}=\mathrm{id}$. This cannot happen as $\varphi_{21}$ must have a kernel. – s.harp Jul 23 at 7:20
• A scenario where you can always extend is if $pBp$ is an ideal in $B$. This would be implied for example by $pB=Bp$, ie $p$ normalising $B$. – s.harp Jul 23 at 7:25