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It is easy to see that for $S^2$ this space is nothing but a circle that is the intersection of a cone with aperture $2\alpha$ (where $\alpha$ is the predifined specific angle), and $S^2$. My question is that is this observation extendable to higher dimensions? Is it true that for any given vector $u\in\mathbb{R}^n$ and the unit vectors in $\mathbb{R}^n$, the geometry of the points over this unit sphere that have a specific angle with $u$ is $S^{n-1}$? Any hints to prove disprove it.

DISCLAIMER: This is not a homework but rather a self observation (as it might be guessed from its silliness(?!))

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In any dimension, this is a sphere. Suppose WLOG that $u=(1,0\ldots,0)$. The condition that a vector makes angle $\alpha$ with you is that the dot product is $\cos\alpha$ (this is essentially the definition of the angle). If $x=(x_1,\ldots,x_n)$ this condition simply means that $x_1=\cos\alpha$. Therefore the set we are talking about is defined by the equation $$x_2^2+\ldots+x_n^2=1-\cos^2\alpha,$$ a sphere of dimension $n-2$.

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General spherical coordinates are given by $$ \begin{align*} x_1 &=r\sin{\phi_1}\sin{\phi_2}\dotsm\sin{\phi_{n-2}}\sin{\phi_{n-1}} \\ x_2 &=r\sin{\phi_1}\sin{\phi_2}\dotsm\sin{\phi_{n-2}}\cos{\phi_{n-1}} \\ x_3 &=r\sin{\phi_1}\sin{\phi_2}\dotsm\cos{\phi_{n-2}} \\ &\vdots\\ x_{n-1} &=r\sin{\phi_1}\cos{\phi_2}\\ x_n &= r \cos{\phi_1} \end{align*}$$ Setting $r=1$ and dotting this with $$ (0,\dotsc,0,1) $$ gives $\cos{\phi_1}$, so if we hold that constant, we have a sphere (looking at the similarity of the coordinates after setting $\phi_1$ constant), and we can see, again by the definition, that it has radius $\lvert \sin{\phi_1} \rvert$ (since the first $n-1$ coordinates have common factor $\sin^2{\phi_1}$, and by analogy with $r$ in the full space).

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