4 component spinor isomorphic to S^7?


by Spinnor
Tags: component, isomorphic, spinor
Spinnor
Spinnor is offline
#1
May7-12, 11:28 AM
P: 1,362
I was told the space S^3 is isomorphic to the set of all 2 component spinors with norm 1 (see http://www.physicsforums.com/showthread.php?t=603404 ). Can I infer that the space of all 4 component spinors with norm 1 is isomorphic to S^7?

If so is a Dirac spinor isomorphic to S^7?

Thanks for any help!
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Ben Niehoff
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#2
May8-12, 01:20 AM
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"4 component spinor" is not specific enough.

2-component spinors transform under Spin(3) which is isomorphic to SU(2), hence S^3. Dirac spinors transform under a reducible rep of Spin(3,1), which is going to be some non-compact space, not a sphere. But there are other 4-component spinors, such as those in Spin(4), Spin(5), or Spin(4,1). None of these are topologically S^7, though.

S^7 is the set of unit octonions, which don't have a group structure (due to the failure of associativity).
samalkhaiat
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#3
May9-12, 02:05 PM
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Quote Quote by Spinnor View Post
I was told the space S^3 is isomorphic to the set of all 2 component spinors with norm 1 (see http://www.physicsforums.com/showthread.php?t=603404 ). Can I infer that the space of all 4 component spinors with norm 1 is isomorphic to S^7?

If so is a Dirac spinor isomorphic to S^7?

Thanks for any help!
Topologically, the manifold defined by [itex]\bar{\Psi}\Psi=1[/itex] is [itex]S^{3}\times \mathbb{R}^{4}[/itex].

Sam

Spinnor
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#4
May10-12, 08:36 AM
P: 1,362

4 component spinor isomorphic to S^7?


Thanks to both of you, Ben and Sam, for clearing that up!

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