TITLE: Ultraweak topology on B(X): Is the map X\otimes X* -> B(X)* isometric?
QUESTION [16 upvotes]: 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:

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

REPLY [6 votes]: I think the OP’s point (1) is not correct; i.e., $B(Z)$ need not be weak$^*$ dense in $B(Z, Z^{**}) (= B(Z^*)$.  You don’t need a Pisier space to show this.   Instead, take a separable reflexive space $X$ that fails the approximation property (AP) (reflexivity is not so important, but it makes the notation and reasoning slightly simpler). As in my first answer, let $(E_n)$ be an increasing sequence of subspaces of $X$ with $E_n$ having dimension $n$ so that $\cup_{n} E_n$ is dense in $X$, let $Y$ be the $\ell_1$ sum of 
$(E_n)$, and let $Q: Y\to X$ be the natural quotient map, defined by 
$Q(e_n)_n : = \sum _n e_n$. As was hinted at in my first answer, $Y^* = W^\perp  + Q^*X$  is a direct sum decomposition of $Y^*$ with the projection onto $Q^*X$ having norm one and where $W$ is the  kernel of $Q$; i.e., $W $ is the subspace of all $(e_n)$  in $Y$ so that  in $X$ $\sum_n e_n = 0$.  
Since $X$ fails the AP, by Grothendieck’s duality theory there is a tensor $T$ in $X^* \hat{\otimes} X$,  $T = \sum_n x_n^* \otimes x_n$ with $x_n^* \in X^*$, $x_n \in X$, $\sum_n \|x_n^*\| \cdot \|x_n\| < \infty$, and for all finite rank operators $S$ on $X$ we have $\langle S, T\rangle = 
\sum_n  \langle x_n^*, Sx_n \rangle =0$ but $\langle I_X, T\rangle = 
\sum_n  \langle x_n^*, x_n \rangle =1$.  Consider now the tensor $T_1$ in $Y^* \hat{\otimes} X$ defined by $T_1 := \sum_n Q^* x_n^* \otimes x_n$. If $J$ denotes the natural isometric embedding from $X$ into $Y^{**} $ induced by $Q$, we see that $\langle T_1, J \rangle = \sum_n \langle Q^* x_n^*, Jx_n \rangle = \sum_n  \langle x_n^*, x_n \rangle =1$. Now consider any $S$ in $B(X,Y)$.  The operator $S$ is compact because $X$ is reflexive and $Y$ has the Schur property. We claim that $\langle T_1, S\rangle $, which is $\sum_n \langle Q^* x_n^*, Sx_n \rangle$, is zero.  If not, since $\|  P_kS-S\| \to 0$ as $k \to \infty$ by compactness of $S$ (here $P_k$ is the natural norm one projection from $Y= (\sum_{n=1}^\infty E_n)_1$ onto $(\sum_{n=1}^k E_n)_1$)      we can choose $k$ s.t.  $\langle  T_1, P_k S\rangle \not=  0$.  But 
$$
\langle  T_1, P_kS\rangle  =  
\sum_n \langle Q^* x_n^*, P_k Sx_n \rangle =
\sum_n \langle  x_n^*, QP_k Sx_n =0
$$
because $QP_k S$ has finite rank.
To see that (1) is false, consider the space $Z:= Y\oplus X$.