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Math Help - Hilbert A-module

  1. #1
    Member Mauritzvdworm's Avatar
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    Hilbert A-module

    Suppose X is a Hilbert A-module. I would like to show that for each x\in X there is a unique y\in X such that x=y\cdot\langle y,y\rangle

    The plan is to use adjoinable operators for each x\in X
    D_x:X\rightarrow A,D_x(y):=\langle x,y\rangle
    L_x:A\rightarrow X,L_x(a)=x\cdot a

    where we take A_A to be a right Hilbert A-module with the inner product defined by \langle a,b\rangle:=a^*b

    I have shown that the operators D_x,L_x are the adjoint of each other
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by Mauritzvdworm View Post
    Suppose X is a Hilbert A-module. I would like to show that for each x\in X there is a unique y\in X such that x=y\cdot\langle y,y\rangle

    The plan is to use adjoinable operators for each x\in X
    D_x:X\rightarrow A,D_x(y):=\langle x,y\rangle
    L_x:A\rightarrow X,L_x(a)=x\cdot a

    where we take A_A to be a right Hilbert A-module with the inner product defined by \langle a,b\rangle:=a^*b

    I have shown that the operators D_x,L_x are the adjoint of each other
    It's true that the operators D_x,L_x are adjoints of each other, but I don't see how to use that to find y, given x.

    To me this looks a bit similar to the polar decomposition of an element of a von Neumann algebra, as done in Gert Pedersen's book C*-algebras and their automorphism groups. In fact, if y = x\cdot\langle x,x\rangle^{-1/3} then x=y\cdot\langle y,y\rangle. Of course, that construction only works if \langle x,x\rangle is invertible. But you can avoid that snag as follows. First note that we may assume that A is unital (if not, then adjoin an identity e to it and extend the action of A on X to an action of the unitised algebra on X). Now define a sequence of elements y_n = x\cdot\bigl(\frac1ne + \langle x,x\rangle^{1/3}\bigr)^{-1} and prove, (as in Pedersen's Proposition 2.2.9) that (y_n) is Cauchy.

    I don't see how to prove the uniqueness part of the problem. If y\cdot\langle y,y\rangle = z\cdot\langle z,z\rangle then \bigl\langle y\cdot\langle y,y\rangle,y\cdot\langle y,y\rangle \bigr\rangle = \bigl\langle z\cdot\langle z,z\rangle,z\cdot\langle z,z\rangle\bigr\rangle, from which you get \langle y,y\rangle^3 = \langle z,z\rangle^3 and hence \langle y,y\rangle = \langle z,z\rangle. But that's as far as I can get.
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  3. #3
    Member Mauritzvdworm's Avatar
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    I think I have found a way to handle to problem, but it is somewhat involved. Here I will give the idea

    Let us define the following operators
    T\in \mathcal{K}(X,A_A)
    S\in\mathcal{K}(A_A,X)
    R\in\mathcal{K}(X) and let a\in A_A
    We can now use these to define an operator on the direct sum Hilbert A-module A\oplus X in the following way
    \[ \left( \begin{array}{cc}<br />
a & T  \\<br />
S & R  \end{array} \right)<br />
\left( <br />
\begin{array}{c}<br />
b\\<br />
x<br />
\end{array}<br />
\right)=\left(<br />
\begin{array}{c}<br />
 ab+Tx\\<br />
Sb+Rx<br />
\end{array}\right)<br />
\]

    This operator clearly is adjoinable (just take the complex conjugate en transpose of the matrix). We can now also show that the operator defined above is in \mathcal{K}(A\oplus X)

    This can be done using embeddings of the form
    a\mapsto \left(\begin{array}{cc}a & 0  \\<br />
0 & 0  \end{array} \right)
    T\mapsto \left(\begin{array}{cc}0 & T  \\<br />
0 & 0  \end{array} \right)
    and similarly for the others.

    now for each x\in X, \left(\begin{array}{cc}0 & D_x  \\<br />
L_x & 0  \end{array} \right) is selfs adjoint element in \mathcal{K}(A\oplus X) which anticommutes with  \left(\begin{array}{cc}1 & 0  \\<br />
0 & -1  \end{array} \right)

    We now exploit the functional calculus together with the function f(x)=x^{\frac{1}{3}} and find that f\left(\begin{array}{cc}0 & D_x  \\<br />
L_x & 0  \end{array} \right)\in\mathcal{K}(A\oplus X) and anticommutes with  \left(\begin{array}{cc}1 & 0  \\<br />
0 & -1  \end{array} \right)

    We can also show that every self adjoint element in \mathcal{K}(A\oplus X) is of the form \left(\begin{array}{cc}0 & D_x  \\<br />
L_x & 0  \end{array} \right)
    (The proof of this is also somewhat involved and for practical reasons I will omit it here)

    Now we have
    \left(\begin{array}{cc}0 & D_x  \\<br />
L_x & 0  \end{array} \right)=\left(\begin{array}{cc}0 & D_y  \\<br />
L_y & 0  \end{array} \right)^3 for some y\in X

    Now we just multiply out the matrices and solve for the bottom left corner which will yield
    L_x=L_yD_yL_y then we have
    L_x(a)=L_yD_yL_y(a)=L_yD_y(y\cdot a)=L_y(\langle y,y\rangle \cdot a)=y\cdot\langle y,y\rangle \cdot a

    so finally we get x=y\cdot\langle y,y\rangle
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  4. #4
    MHF Contributor
    Opalg's Avatar
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    That's a very nice argument. I haven't checked every detail of it, but it looks convincing. Certainly the cube root function has to lie at the heart of the proof.
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