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?

  • $\begingroup$ $(a+b)^2=a^2+2ab+b^2$? $\endgroup$ Dec 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 Weaver
    Dec 31, 2014 at 1:13

1 Answer 1


Yes. The theorem was proven independently by Blecher, A new approach to C*-modules, Math. Ann. 307 (1997), 253-290 and Lance, Unitary operators on Hilbert C*-modules, Bull. London Math. Soc. 26 (1994), 363-366. (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.


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