Is the set of points in $\mathbb R^n$ with $\sum_{j=1}^n x_j^k = 0$ a submanifold?

Question

Consider the set

$$A:= \{x\in \mathbb R^n :\sum_{j=1}^n x_j^k = 0\}$$

for $k$ an odd integer. Is this a submanifold of $\mathbb R^n$ for every $n$? For $n=1$, it is just 0; for $n = 2$, it is the anti-diagonal $\{(x_1 , -x_1) : x_1 \in \mathbb R\}$, which is a submanifold. However, I cannot find a way to determine this in higher dimensions. Any suggestions?

Preimage by some real function? What can you say about preimages? — Sigur, Jan 23 '13 at 00:11
What's the role of $k$ here? For $k = n = 2$, we have $A = {(0,0)}$. — Adam Saltz, Jan 23 '13 at 00:11
Well, typically we can say that a pre-image of a regular value is a submanifold. However, here 0 is not a regular value of $x\mapsto \sum_j x_j^k$ unless $k = 1$. — user15464, Jan 23 '13 at 00:21
@user15464: Remove the origin from your domain, and then $0$ is a regular value. — , Jan 23 '13 at 10:53

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

Your set is scale-invariant: if $x\in A$ and $t\in\mathbb R$ then $tx\in A$. If such a set is a submanifold, then it is a linear subspace. If $n>2$, $A$ is not a linear subspace, hence not a manifold.

edited Apr 13 '17 at 12:20

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answered Jan 23 '13 at 01:30

Is the set of points in $\mathbb R^n$ with $\sum_{j=1}^n x_j^k = 0$ a submanifold?

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