Regarding the complete reducability of the Lorentz group
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I was just reading that the Lorentz group has the property of complete reducability, that is any representation can be written as the direct sum of irreducible representations.
This reminds me somewhat of the Peter weyl theorem, in that representation of $G$ on $L^{2}(G)$ (for compact $G$) may be written as a sum of irreducible unitary representations (the matrix coefficients of which form an orthonormal basis).
So I know the Lorentz group is definitely non-compact; however I'm curious as to whether the irreducible representations form an orthonormal basis. If not, How does one go about expressing a general representation in terms of a sum of irreducible representations?
lie-groups semi-riemannian-geometry
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add a comment |
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I was just reading that the Lorentz group has the property of complete reducability, that is any representation can be written as the direct sum of irreducible representations.
This reminds me somewhat of the Peter weyl theorem, in that representation of $G$ on $L^{2}(G)$ (for compact $G$) may be written as a sum of irreducible unitary representations (the matrix coefficients of which form an orthonormal basis).
So I know the Lorentz group is definitely non-compact; however I'm curious as to whether the irreducible representations form an orthonormal basis. If not, How does one go about expressing a general representation in terms of a sum of irreducible representations?
lie-groups semi-riemannian-geometry
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This should follow from Weyl's theorem of complete reducibility: en.wikipedia.org/wiki/Weyl%27s_theorem_on_complete_reducibility
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– Max
Dec 13 '18 at 3:35
add a comment |
$begingroup$
I was just reading that the Lorentz group has the property of complete reducability, that is any representation can be written as the direct sum of irreducible representations.
This reminds me somewhat of the Peter weyl theorem, in that representation of $G$ on $L^{2}(G)$ (for compact $G$) may be written as a sum of irreducible unitary representations (the matrix coefficients of which form an orthonormal basis).
So I know the Lorentz group is definitely non-compact; however I'm curious as to whether the irreducible representations form an orthonormal basis. If not, How does one go about expressing a general representation in terms of a sum of irreducible representations?
lie-groups semi-riemannian-geometry
$endgroup$
I was just reading that the Lorentz group has the property of complete reducability, that is any representation can be written as the direct sum of irreducible representations.
This reminds me somewhat of the Peter weyl theorem, in that representation of $G$ on $L^{2}(G)$ (for compact $G$) may be written as a sum of irreducible unitary representations (the matrix coefficients of which form an orthonormal basis).
So I know the Lorentz group is definitely non-compact; however I'm curious as to whether the irreducible representations form an orthonormal basis. If not, How does one go about expressing a general representation in terms of a sum of irreducible representations?
lie-groups semi-riemannian-geometry
lie-groups semi-riemannian-geometry
asked Dec 13 '18 at 3:18
R. RankinR. Rankin
333213
333213
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This should follow from Weyl's theorem of complete reducibility: en.wikipedia.org/wiki/Weyl%27s_theorem_on_complete_reducibility
$endgroup$
– Max
Dec 13 '18 at 3:35
add a comment |
$begingroup$
This should follow from Weyl's theorem of complete reducibility: en.wikipedia.org/wiki/Weyl%27s_theorem_on_complete_reducibility
$endgroup$
– Max
Dec 13 '18 at 3:35
$begingroup$
This should follow from Weyl's theorem of complete reducibility: en.wikipedia.org/wiki/Weyl%27s_theorem_on_complete_reducibility
$endgroup$
– Max
Dec 13 '18 at 3:35
$begingroup$
This should follow from Weyl's theorem of complete reducibility: en.wikipedia.org/wiki/Weyl%27s_theorem_on_complete_reducibility
$endgroup$
– Max
Dec 13 '18 at 3:35
add a comment |
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This should follow from Weyl's theorem of complete reducibility: en.wikipedia.org/wiki/Weyl%27s_theorem_on_complete_reducibility
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– Max
Dec 13 '18 at 3:35