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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?

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Preimage by some real function? What can you say about preimages? –  Sigur Jan 23 '13 at 0:11
    
What's the role of $k$ here? For $k = n = 2$, we have $A = \{(0,0)\}$. –  Adam Saltz Jan 23 '13 at 0:11
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@Adam: $k$ is odd. –  Jason DeVito Jan 23 '13 at 0:19
    
@JasonDeVito Thank you. –  Adam Saltz Jan 23 '13 at 0:20
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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 0:21

1 Answer 1

up vote 4 down vote accepted

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.

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