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Does the Hairy Ball theorem imply the Borsuk-Ulam for even dimensions?
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Consider the Hairy Ball theorem, stating that a continuous vector field over $R^2n$ must have a point where the tangential component vanishes,
and the Borsuk-Ulam theorem, stating that any continuous function $f: S^k to mathbbR^k$ must send two antipodal points to the same image.
I was wondering if, for $k = 2n$, the Borsuk-Ulam theorem follows from the Hairy Ball theorem.
The idea would be to take a continuous function $f: S^2n to mathbbR^2n$ and define $g: S^2n to mathbbR^2n$ as $g(P) = f(P) - f(-P)$. If we think of $g$ as a tangential vector field on $S^2n$, it follows that $exists P in S^2n: g(P) = f(P) - f(-P) = 0$ and the Borsuk-Ulam theorem (for even dimensions) follows.
I am not sure if I can "think of $g$ as a tangential vector field on $S^2n$", otherwise I believe the proof is fine.
Any thoughts on this?
proof-verification algebraic-topology vector-fields
asked Jul 31 at 16:39
RGS
8,19711129
add a comment |Â
up vote
2
down vote
favorite
Consider the Hairy Ball theorem, stating that a continuous vector field over $R^2n$ must have a point where the tangential component vanishes,
and the Borsuk-Ulam theorem, stating that any continuous function $f: S^k to mathbbR^k$ must send two antipodal points to the same image.
I was wondering if, for $k = 2n$, the Borsuk-Ulam theorem follows from the Hairy Ball theorem.
The idea would be to take a continuous function $f: S^2n to mathbbR^2n$ and define $g: S^2n to mathbbR^2n$ as $g(P) = f(P) - f(-P)$. If we think of $g$ as a tangential vector field on $S^2n$, it follows that $exists P in S^2n: g(P) = f(P) - f(-P) = 0$ and the Borsuk-Ulam theorem (for even dimensions) follows.
I am not sure if I can "think of $g$ as a tangential vector field on $S^2n$", otherwise I believe the proof is fine.
Any thoughts on this?
proof-verification algebraic-topology vector-fields
asked Jul 31 at 16:39
RGS
8,19711129
I agree: I cannot see how to think of $g$ as a tangential vector field on $S^2n$.
– Lord Shark the Unknown
Jul 31 at 17:11
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the Hairy Ball theorem, stating that a continuous vector field over $R^2n$ must have a point where the tangential component vanishes,
and the Borsuk-Ulam theorem, stating that any continuous function $f: S^k to mathbbR^k$ must send two antipodal points to the same image.
I was wondering if, for $k = 2n$, the Borsuk-Ulam theorem follows from the Hairy Ball theorem.
The idea would be to take a continuous function $f: S^2n to mathbbR^2n$ and define $g: S^2n to mathbbR^2n$ as $g(P) = f(P) - f(-P)$. If we think of $g$ as a tangential vector field on $S^2n$, it follows that $exists P in S^2n: g(P) = f(P) - f(-P) = 0$ and the Borsuk-Ulam theorem (for even dimensions) follows.
I am not sure if I can "think of $g$ as a tangential vector field on $S^2n$", otherwise I believe the proof is fine.
Any thoughts on this?
proof-verification algebraic-topology vector-fields
asked Jul 31 at 16:39
RGS
8,19711129
Consider the Hairy Ball theorem, stating that a continuous vector field over $R^2n$ must have a point where the tangential component vanishes,
and the Borsuk-Ulam theorem, stating that any continuous function $f: S^k to mathbbR^k$ must send two antipodal points to the same image.
I was wondering if, for $k = 2n$, the Borsuk-Ulam theorem follows from the Hairy Ball theorem.
The idea would be to take a continuous function $f: S^2n to mathbbR^2n$ and define $g: S^2n to mathbbR^2n$ as $g(P) = f(P) - f(-P)$. If we think of $g$ as a tangential vector field on $S^2n$, it follows that $exists P in S^2n: g(P) = f(P) - f(-P) = 0$ and the Borsuk-Ulam theorem (for even dimensions) follows.
I am not sure if I can "think of $g$ as a tangential vector field on $S^2n$", otherwise I believe the proof is fine.
Any thoughts on this?
proof-verification algebraic-topology vector-fields
asked Jul 31 at 16:39
RGS
8,19711129
asked Jul 31 at 16:39
RGS
8,19711129
asked Jul 31 at 16:39
RGS
8,19711129
asked Jul 31 at 16:39
RGS
8,19711129
8,19711129
I agree: I cannot see how to think of $g$ as a tangential vector field on $S^2n$.
– Lord Shark the Unknown
Jul 31 at 17:11
add a comment |Â
I agree: I cannot see how to think of $g$ as a tangential vector field on $S^2n$.
– Lord Shark the Unknown
Jul 31 at 17:11
I agree: I cannot see how to think of $g$ as a tangential vector field on $S^2n$.
– Lord Shark the Unknown
Jul 31 at 17:11
I agree: I cannot see how to think of $g$ as a tangential vector field on $S^2n$.
– Lord Shark the Unknown
Jul 31 at 17:11
add a comment |Â
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I agree: I cannot see how to think of $g$ as a tangential vector field on $S^2n$.
– Lord Shark the Unknown
Jul 31 at 17:11