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Consider some Hilbert space $(\mathcal{H},\Vert\cdot\Vert)$. Now, there are several equivalent definitions of unitary operator, i.e. see for example on wikipedia. All of them assume either that $U$ presevers the inner product or that $U$ commutes with its adjoint, but then also some additional stuff, like surjectivity or boundedness. However, doesn't these further properties not already follow from the fact that $UU^{\ast}=U^{\ast}U=\mathrm{id}_{\mathcal{H}}$?

Only using the fact that $UU^{\ast}=U^{\ast}U=\mathrm{id}_{\mathcal{H}}$, we can deduce the following properties:

  1. $U$ is bijective with inverse $U^{-1}=U^{\ast}$, by definition, since $UU^{\ast}=U^{\ast}U=\mathrm{id}_{\mathcal{H}}$ exactly tells us that $U^{\ast}$ is the left and right inverse of $U$.
  2. $U$ preserves the inner produce, since $\langle Ux,Uy\rangle=\langle x,U^{\ast}Uy\rangle=\langle x,y\rangle$ for all $x,y\in\mathcal{H}$, just by defintion of the adjoint.
  3. $U$ is linear, since \begin{aligned}\|\lambda U(x)-U(\lambda x)\|^{2}&=\langle \lambda U(x)-U(\lambda x),\lambda U(x)-U(\lambda x)\rangle \\&=\|\lambda U(x)\|^{2}+\|U(\lambda x)\|^{2}-\langle U(\lambda x),\lambda U(x)\rangle -\langle \lambda U(x),U(\lambda x)\rangle \\&=|\lambda |^{2}\|U(x)\|^{2}+\|U(\lambda x)\|^{2}-{\overline {\lambda }}\langle U(\lambda x),U(x)\rangle -\lambda \langle U(x),U(\lambda x)\rangle \\&=|\lambda |^{2}\|x\|^{2}+\|\lambda x\|^{2}-{\overline {\lambda }}\langle \lambda x,x\rangle -\lambda \langle x,\lambda x\rangle \\&=0\end{aligned} and similarely $\| U(x+y)-(Ux+Uy)\| = 0$ and hence $\lambda U(x)=U(\lambda x)$ and $U(x+y)=U(x)+U(y)$ by postive-definiteness of $\Vert\cdot\Vert$.
  4. $U$ is bounded, since $\Vert Ux\Vert = \Vert x\Vert$ for all $x\in\mathcal{H}$ and hence $$\Vert U\Vert=\sup_{x\in\mathcal{H},\Vert x\Vert=1}\Vert Ux\Vert=\sup_{x\in\mathcal{H},\Vert x\Vert=1}\Vert x\Vert=1<\infty$$

For example, the first definition on wikipedia is the following:

"A unitary operator is a bounded linear operator $U:\mathcal{H}\to\mathcal{H}$ on a Hilbert space $\mathcal{H}$ that satisfies $U^{\ast}U=UU^{\ast}=\mathrm{id}_{\mathcal{H}}$"

So they assume in addition boundedness. However, does this not already follow from the other property, as explained in my point 4. above. As another example, wikipedia states the following equivalent definition:

"A unitary operator $U:\mathcal{H}\to\mathcal{H}$ is a surjective linear operator on some Hilbert space $\mathcal{H}$ that presevres the inner product"

But again, using the fact that $U$ preserves the norm, it follows that $U^{-1}=U^{\ast}$, which means that $U$ is bijective and hence also surjective.

Where is my thinking error?

  • How does $U^{-1}=U^*$ follow from $U$ preserving the norm? What if $U$ is not invertible at all? – Eric Wofsey Dec 22 '21 at 20:17
  • Sorry that was a typo. It should mean "preserves inner product" –  Dec 22 '21 at 20:23
  • If $U$ preserves the inner product we have that $\langle x,y\rangle=\langle Ux,Uy\rangle=\langle x,U^{\ast}Uy\rangle$ and hence $\langle x,U^{\ast}Uy-y\rangle$. Since this is true for all $x$, we conclude that $U^{\ast}Uy-y=0$ for all $y$ by non-degeneracy of the inner product. Hence we get that $U^{\ast}U=\mathrm{id}$, i.e. $U$ admits a left-inverse, namely $U^{\ast}$. Similarely we show that $UU^{\ast}=\mathrm{id}$ and hence, $U$ is invertible with inverse $U^{\ast}$. –  Dec 22 '21 at 20:25
  • I agree that $U^U=id$ but how do you "similarly" get $UU^=id$? – Eric Wofsey Dec 22 '21 at 20:26
  • Okay, I see thats actually a good point^^ –  Dec 22 '21 at 20:28
  • So, is this the reason why we need surjectivity in the second definition? But what about the first definition? Is boundedness as well as linearity redundant as I wrote above? –  Dec 22 '21 at 20:30

1 Answers1

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The condition $U^*U=I=UU^*$ is exactly equivalent to "$U$ is a surjective isometry". In particular, as you say, as an isometry $U$ is bounded and the conditions also force linearity.

The problem with removing "linear and bounded" from the definition is that you you cannot guarantee that $U^*$ exists.

Not sure if it helps, but the morphisms of Hilbert spaces are the isometries, and the isomorphisms are the unitaries. Here I mean to emphasize how natural they are.

Martin Argerami
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  • The meaning of U* is unclear without further explanation if you don't assume that U is linear. – Martti Karvonen Dec 23 '21 at 08:59
  • How? $U^$ is the function such that $$\langle Ux,y\rangle=\langle x,U^y\rangle$$ for all $x,y$. There's no linearity there. What is true, and I'll edit that in, is that one cannot know it exists in general. – Martin Argerami Dec 23 '21 at 14:45
  • I agree, it's just that the existence and uniqueness of U* are not obvious without linearity. Hence speaking of U* as "the function" is misleading, as there could be many, or none. – Martti Karvonen Dec 23 '21 at 15:28
  • You don't need linearity for the uniqueness. If $$\langle x,f(y)\rangle=\langle x,g(y)\rangle$$ for all $x$, then $f(y)=g(y)$. Existence is the issue. Which shows, in my view, that surjective isometry is the best definition, as it implies everything else. – Martin Argerami Dec 23 '21 at 15:54
  • You're right, it's just the existence that's at stake. – Martti Karvonen Dec 23 '21 at 16:07