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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 $\phi':B\rightarrow\mathcal{B}(H)$?

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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.

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  • $\begingroup$ Thanks for your answer. I wonder if the answer would be different if we only allow $H$ to be infinite-dimensional? $\endgroup$ – ougoah Jul 23 at 1:24
  • $\begingroup$ 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. $\endgroup$ – s.harp Jul 23 at 7:20
  • $\begingroup$ 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 $B$. $\endgroup$ – s.harp Jul 23 at 7:25

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