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If I have two Hilbert spaces $X$ and $Y$ and a continuous linear isomorphism $T:X \to Y$ with continuous inverse $T^{-1}:Y\to X$, is there anyway to write $$(a,b)_X$$ as an inner product on $Y$?

I know I can do this if $T$ is an isometric isomorphism, but if it's not, is there anything I can do?

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The pullback $(T^{-1})^\ast(\cdot,\cdot)_X$ of $(\cdot,\cdot)_X$ from $X$ to $Y$ via $T^{-1}$, defined by $$(T^{-1})^\ast(\eta,\xi)_X := (T^{-1}\eta,T^{-1}\xi)_X, \quad \eta,\xi \in Y,$$ will be some inner product on the vector space $Y$ inducing a norm equivalent to the original Hilbert space norm on $Y$, whether or not $T$ is isometric. In particular, then, $(T^{-1})^\ast(\cdot,\cdot)_X = (\cdot,\cdot)_Y$ if and only if $T^{-1}$ is an isometry, if and only if $T$ is an isometry.

EDIT: In particular, $$(T^{-1})^\ast(\eta,\xi)_X := (T^{-1}\eta,T^{-1}\xi)_X = (\eta,(T T^\ast)^{-1}\xi)_Y, \quad \eta,\xi \in Y,$$ so that understanding $(T^{-1})^\ast(\cdot,\cdot)_X$ is the same as understanding the operator $TT^\ast$ on $Y$ and its inverse $(TT^\ast)^{-1}$.

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Thanks. But one can't write down a formula for the inner product on $Y$ explicitly, I guess :( –  george.s Jan 22 '13 at 23:12
    
@george.s See the edit for the most that can be said a priori. Basically, the best you can do, given what you have, is to compare $(T^{-1})^\ast(\cdot,\cdot)_X$ with $(\cdot,\cdot)_Y$ as inner products on $Y$, and by the edit, understanding $(T^{-1})^\ast(\cdot,\cdot)_X$ in terms of $(\cdot,\cdot)_Y$ is equivalent to understanding the operator $(TT^\ast)^{-1}$ on $Y$ with $(\cdot,\cdot)_Y$. –  Branimir Ćaćić Jan 23 '13 at 1:47
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