In Hilbert space theory if we have a linear bijection $U:H \to K$ which is an isometry, then it preserves inner products. Suppose now that you have two (right) Hilbert modules $X,Y$ (say, over the same $C^*$algebra $B$) and a $B$linear isomorphism $T:X \to Y$ which is an isometry (recall that the norm in Hilbert module is defined via $\x\^2_X:=\ \langle x,x\rangle\_B$). Does it follows that $T$ preserves $B$valued inner products?

$\begingroup$ $(a+b)^2=a^2+2ab+b^2$? $\endgroup$– Alex DegtyarevDec 30, 2014 at 20:49

$\begingroup$ I take it Alex is coyly referring to a polarization identity. Nik's answer seems more adequate. $\endgroup$– Todd Trimble ♦Dec 31, 2014 at 0:40

1$\begingroup$ @Todd  the subtlety is that we don't know $\langle x,x\rangle = \langle Tx, Tx\rangle$  then Alex's answer would work  we only know $\\langle x,x\rangle\_B = \\langle Tx, Tx\rangle\_B$. The "inner product" on a Hilbert module takes values in a C*algebra, but the norm is still a scalar. $\endgroup$– Nik WeaverDec 31, 2014 at 1:13
1 Answer
Yes. The theorem was proven independently by Blecher, A new approach to C*modules, Math. Ann. 307 (1997), 253290 and Lance, Unitary operators on Hilbert C*modules, Bull. London Math. Soc. 26 (1994), 363366. (Google would have told you this.) [Correction: the result is due to Lance. However, the formula below which recovers the inner product is Blecher's.]
In fact, the inner product can be recovered from the norm and $B$module structure by the formula $\langle x,x\rangle = \sup\{r(x)^*r(x): r: X \to B$ is $B$linear and $\r\ \leq 1\}$, and then you get $\langle x,y\rangle$ by polarization.