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Let $A$ be a real square matrix. The condition that $A$ preserves the inner product, and hence preserves lengths, is the interpretation of $A \in O(n)$. If we further require that $det(A) = 1$, then the action of $A$ is strictly a rotation and not a reflection.

I would like to obtain a similar understanding for the geometric actions of $U(n)$ and $SU(n)$. Let $A$ be a complex square matrix. The group $U(n)$ is the complex analogue of $O(n)$, meaning that $A$ preserves the complex inner product and again similarly preserves lengths in $\mathbb{C}^n$.

An arbitrary matrix $A \in U(n)$ has determinant $e^{i\theta}$ for some $\theta \in [-\pi, \pi]$. I am having trouble interpreting what this means geometrically. Intuitively it seems like some sort of "continuous reflection" if such a thing exists. How can we interpret the geometric meaning of the condition $det(A) = 1$ in $SU(n)$?

A similar question was posted (How to Interpret $SU(n)$ Geometrically?) where a comment says

An element of $SU(n)$ is naturally a rotation of $\mathbb{R}^{2n}$, however it's not an arbitrary one. They are specifically those rotations which behave well with the additional structure on $\mathbb{C}^n$ given by multiplying by a complex number.

What is meant by "behaves well with the additional structure on $\mathbb{C}^n$?

CBBAM
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    I think it is a strange way to say that elements of $SU(n),$ complex matrices, are $\mathbb{C}$-linear. The author first described them as real rotations and thus lost $\mathbb{C}$-linearity. The statement tried to "save" it without turning back to complex spaces. – Marius S.L. Aug 11 '23 at 20:27
  • @MariusS.L. Thank you for your comment. So an element in $SU(n)$ can be interpreted as a rotation in $\mathbb{R}^{2n}$ that is algebraically "nice" with respect to $\mathbb{C}$? Is there any interpretation that does not use the identification $\mathbb{R}^{2n} \cong \mathbb{C}^n$? – CBBAM Aug 11 '23 at 20:32
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    Yes. My first thought was that it does not change the volume if (complex) variables are transformed by $SU(n).$ This is generally true for matrices of determinant one. Here is an article about $SU(2)$ https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ . It is easier to "see" the geometry in low dimensions (Hopf fibration.) – Marius S.L. Aug 11 '23 at 20:43
  • @MariusS.L. I am currently learning about the Hopf fibration so this is perfect, thank you for sharing. – CBBAM Aug 11 '23 at 20:48
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    @MariusS.L. That's true for $U(n)$ as well. A linear transformation will preserve volume if and only if its determinant has absolute value (or "modulus" if you prefer) equal to $1$, which holds for all unitary matrices. – Ben Grossmann Aug 12 '23 at 16:02

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