The adjoint operator

For $T\in\mathit{CB}(X,Y)$, the adjoint operator $T^*$ is defined as usual. We have: $T^*\in \mathit{CB}(Y^*,X^*)$, and $\Vert T\Vert _{\mathrm{cb}}=\Vert T^*\Vert _{\mathrm{cb}}$. The mapping

$${}^*:\mathit{CB}(X,Y) \to \mathit{CB}(Y^*,X^*)$$

$$T \mapsto T^*$$

even is completely isometric [Ble92b, Lemma 1.1].¹³

$T^*$ is a complete quotient mapping if and only if $T$ is completely isometric; $T^*$ is completely isometric if $T$ is a complete quotient mapping. Especially for a subspace $X_0\subset X$ we have [Ble92a]:

$$X_0^*\stackrel{\mathrm{cb}}{=}X^*/X_0^{\perp}$$

and, if $X_0$ is closed,

$$(X/X_0)^*\stackrel{\mathrm{cb}}{=}X_0^{\perp}$$   .

Footnotes

...Blecher92a.¹³

The isometry on the matrix levels follows from the isometry on the first matrix level using the above mentioned formula $\mathit{CB}({\mathbb{M}}_n(X)^*,Y)\stackrel{\mathrm{cb}}{=}\mathit{CB}(X^*,{\mathbb{M}}_n(Y))$:

$$M_n(\mathit{CB}(X,Y)) = M_1(\mathit{CB}(X,{\mathbb{M}}_n(Y)))$$

$$\hookrightarrow M_1(\mathit{CB}({\mathbb{M}}_n(Y)^*,X^*))$$

$$= M_1(\mathit{CB}(Y^*,{\mathbb{M}}_n(X^*)))$$

$$= M_n(\mathit{CB}(Y^*,X^*))$$