I'm told that $SU(n)$ acts transitively on $S^{2n-1} \subset \mathbb{C}^n$ by matrix multiplication. Yet I can't find a proof of this anywhere, so I was trying to construct a proof on my own by mimicking a version of the proof that I know for showing $SO(n)$ acts transitively on $S^n$.

My proof is outlined as follows:

To show transitive, it suffices to show that the point $x= \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \in S^5 \subset \mathbb{C}^3$ can be taken to any other point $p=\begin{pmatrix} p_1 \\ p_2 \\ p_3 \end{pmatrix} \in S^5 \subset \mathbb{C}^3. $

A complex matrix lies in $SU(3)$ if and only if its columns form an orthonormal basis for $\mathbb{C}^3$.

If $A \in SU(3)$ has the property that $Ax=p$, then $A$ must have $p$ as its first column. Completing $p$ to a basis for $\mathbb{C}^3$, then running the Gram-Schmidt algorithm on this basis (starting with $p$ in the first step of the algorithm) completes $p$ to an orthonormal basis for $\mathbb{C}^3$. Then sticking these basis elements in as the columns of $A$, with $p$ as the first column, yields a matrix in $U(n)$ that takes $x$ to $p$.

This proof in general shows that $U(n)$ acts transitively on $S^{2n-1}$, but the matrix I have constructed does not necessarily have determinant $1$. In the completely analogous proof for showing $SO(n)$ acts on $S^n$ transitively, the matrix has determinant $\pm1$ so if the determinant is $-1$, we need only interchange the last two columns to get a matrix with determinant $1$.

This proof breaks down here because we only know that the determinant lies on the unit circle in $\mathbb{C}$, that is $|\det(A)|=1$. Is there some way to produce a matrix with determinant dead equal to $1$ with this process? Otherwise, is there another simple proof of transitivity?


You can multiply the last column of your matrix with any complex number of absolut value $1$. This does not change the orthonormality of the columns nor the first column (if $n>1$), but multiplies the determinant by that number, so we can certainly make the determinant $=1$.

  • $\begingroup$ Thanks, that sorts it out easily. $\endgroup$ – TuoTuo Jul 6 at 23:36

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