homomorphisms of B(H)

Question

Let $\varphi\colon B(H)\to B(H)$ is an injective $\ast-$homomorphism ($\varphi$-linear, $\varphi(xy)=\varphi(x)\varphi(y)$, $\varphi(x^{\ast})=\varphi(x)^{\ast}$), where $H-$separable Hilbert space. Is it true that $\varphi$ is actually $\ast-$isomorphism? In other words $\varphi$ is also a surjection.

Answer 1

No, not if $H$ is infinite dimensional. In the case that $H$ is finite dimensional, $B(H)$ also is finite dimensional and any linear injection $B(H)\to B(H)$ necessarily is invertible.

For an infinite dimensional Hilbert space you have that $H\cong H\oplus H$. As such the map $B(H)\to B(H\oplus H)$, $A\mapsto A\oplus 0$ can be viewed as a a map $B(H)\to B(H)$. Note that this map is clearly an injective $*$-morphism, but it is not surjective. Further the morphism is not unital, which may make some people sad, but this can be fixed by considering the map $A\mapsto A\oplus A$, which is now a unital injective $*$-morphism that is not surjective.

Answer 2

the map A↦A⊕A, which is a unital injective ∗-morphism that is not

surjective somewhat confusing me. Because, in the book Douglas

"Banach Algebra Techniques in Operator Theory" is proved the

following theorem (Corollary 5.41, page 143): If $A$ is $C^{\ast}$ algebra on Hilbert space $H$ which contains $K(H)-$(the set of compact operators) and $\varphi$ is $\ast-$homomorphism of $A$ into $B(H)$ such that $\varphi|_{K(H)}$ is not zero and $\varphi(A)$ is irreducible, then there exists an unitary operator $u$ on $H$ such that $\varphi(x)=uxu^{\ast}$ for all $x\in A$.

From here we obtain $\varphi$ that is surjection. (A subset $G\subset B(H)$ is said to be irreducible if no proper closed subspace is reducing for all $x$ in $G$).

So, if ∗-morphism $\varphi(A)=A⊕A$ is unital, then $\varphi(e)= e$, where $e$ is identity operator and hence $\varphi(B(H))$ is irreducible. Clearly $\varphi|_{K(H)}$ is not zero. So, could you please tell me where I'm wrong?