Let $X$ be a Banach space. Consider the map $$ \alpha\colon X\hat{\otimes} X^* \to B(X)^*, $$ defined one simple tensors as $$ \alpha(\xi\otimes\eta)(a) = \eta(a(\xi)).\quad (\xi\in X, \eta\in X^*, a\in B(X)) $$ Put differently, we consider the pairing between $X\hat{\otimes} X^*$ (the projective tensor product of $X$ and its dual space $X^*$) and $B(X)$ (the algebra of bounded linear operators on $X$) induced by $$ \langle \xi\otimes\eta, a\rangle = \langle a(\xi), \eta \rangle. $$

Question: Is $\alpha$ isometric?

This question arose when I tried to make sense of the ultraweak topology on $B(X)$ for a (nonreflexive) Banach space $X$. The question asks if $B(X)$ always has a distinguished part of its dual, which could be thought of as a generalization of a predual. However, even if $\alpha$ is isometric, I don't think it necessarily implies that $X\hat{\otimes} X^*$ is a predual of $B(X)$.

One reformulation of the question is as follows: Consider the map $$ \beta\colon B(X) \to B(X^*) $$ which sends an operator $a\in B(X)$ to its transpose $a^*\in B(X^*)$. This map is isometric and we can therefore think of $B(X)$ as a subspace of $B(X^*)$. Then $\alpha$ is isometric if and only if $B(X)$ is $1$-norming in $B(X^*)$ - meaning that the unit ball of $B(X)$ is weak-dense in the unit ball of $B(X^*)$, for the weak-topology on $B(X^*)$ induced by the isometric isomorphism $B(X^*)\cong (X\hat{\otimes} X^*)^*$.

Some more remarks:

1. We know that $B(X)$ is always weak*-dense in $B(X^*)$. (In general, this is weaker then being $1$-norming.) Translating back to $\alpha$, this means that $\alpha$ is always injective.

2. If $X$ is reflexive, then $\beta$ is surjective, and so $\alpha$ is isometric. (This can also be seen more directly.)

3. If $X$ has the metric approximation property, then already the unit ball of $F(X)$, the finite rank operators on $X$, is weak*-dense in the unit ball of $B(X^*)$. Thus, also in this case, $\alpha$ is isometric.

Let $X$ be a Banach space. Consider the map $$ \alpha\colon X\hat{\otimes} X^* \to B(X)^*, $$ defined one simple tensors as $$ \alpha(\xi\otimes\eta)(a) = \eta(a(\xi)).\quad (\xi\in X, \eta\in X^*, a\in B(X)) $$ Put differently, we consider the pairing between $X\hat{\otimes} X^*$ (the projective tensor product of $X$ and its dual space $X^*$) and $B(X)$ (the algebra of bounded linear operators on $X$) induced by $$ \langle \xi\otimes\eta, a\rangle = \langle a(\xi), \eta \rangle. $$

Question: Is $\alpha$ isometric?

This question arose when I tried to make sense of the ultraweak topology on $B(X)$ for a (nonreflexive) Banach space $X$. The question asks if $B(X)$ always has a distinguished part of its dual, which could be thought of as a generalization of a predual. However, even if $\alpha$ is isometric, I don't think it necessarily implies that $X\hat{\otimes} X^*$ is a predual of $B(X)$.

One reformulation of the question is as follows: Consider the map $$ \beta\colon B(X) \to B(X^*) $$ which sends an operator $a\in B(X)$ to its transpose $a^*\in B(X^*)$. This map is isometric and we can therefore think of $B(X)$ as a subspace of $B(X^*)$. Then $\alpha$ is isometric if and only if $B(X)$ is $1$-norming in $B(X^*)$ - meaning that the unit ball of $B(X)$ is weak-dense in the unit ball of $B(X^*)$, for the weak-topology on $B(X^*)$ induced by the isometric isomorphism $B(X^*)\cong (X\hat{\otimes} X^*)^*$.

Some more remarks:

1. Originally I had written: "We know that $B(X)$ is always weak*-dense in $B(X^*)$. (In general, this is weaker then being $1$-norming.) Translating back to $\alpha$, this means that $\alpha$ is always injective." However, as pointed out by Bill Johnson below, not even that is the case. So the answer to my question is a big "NO".

2. If $X$ is reflexive, then $\beta$ is surjective, and so $\alpha$ is isometric. (This can also be seen more directly.)

3. If $X$ has the metric approximation property, then already the unit ball of $F(X)$, the finite rank operators on $X$, is weak*-dense in the unit ball of $B(X^*)$. Thus, also in this case, $\alpha$ is isometric.