- #1
LCSphysicist
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- TL;DR
- Under a special type of coordinate transformation, the KK reduce metric transforms such that a non-abelian gauge vector appears. How?
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The Kaluza-Klein metric, by reduction, can be written as a ##(4+m) \times (4+m)##symmetric matrix, where ##m## is the dimension of the additional spacetime (if we decompose ##M_D = M_4 \times M_m##). It was show by Bryce de. Witt that, if the non-diagonal metric has the form
$$g_{\mu m} = B^I_{\mu}(x^{\nu}) V^I_m(y^n)$$
And we make a general coordinate transformation
$$\epsilon_m = \lambda^I (x^{\nu}) V_m^I(y^n)$$
We can show that ##\lambda## is the parameter of a non-abelian gauge transformation of ##B_{\mu}##. While every source talks about de Witt derivation, i was not able to find it, indeed, the citations used are not open (B.S. DeWitt, In Relativity, Groups and Topology, eds. C. and B.S. DeWitt (Gordon and Breach, New York, 1964). So i have tried to show it here
$$
\delta g_{rs} = \epsilon^m \partial_m g_{rs} + \partial_{r} \epsilon^m g_{m s} + \partial_s \epsilon^m g_{m s}
$$
$r = \mu, s = n$
$$
\delta g_{\mu n} = \lambda V^{m} \partial_m g_{\mu n} + \partial_{\mu} \lambda V^m g_{mn} + \lambda \partial_n V^m g_{m \mu}
$$
$g_{\mu n} = B_{\mu} V_n$, $g_{mn} = \delta_{mn}$
$$
\delta g_{\mu n} = \delta B_{\mu} V_n + B_{\mu} \delta V_{n} = \lambda B_{\mu} V^m \partial_m V_n + V_n \partial_{\mu} \lambda + \lambda B_{\mu} \partial_n V^m V_m
$$
But ##\delta V_n## is ##0## (honestly, i am just assuming it). Also, i assumed that ##\partial_m V_n V^n = V^n \partial_m V_n##
$$
\delta B_{\mu} V_n = \lambda B_{\mu} V^m \partial_m V_n + V_n \partial_{\mu} \lambda + \lambda B_{\mu} V_m \partial_n V^m
$$
This, in fact, reduces to a "looks like" non-gauge transformation
$$
\delta B_{\mu} = \partial_{\mu} \lambda + (V_n^{-1} V_m \partial_m V_n + V_n^{-1} V_m \partial_n V^m ) \lambda B_{\mu}
$$
But i am not really sure if it is correct. I mean, the term in parenthesis shouldn't be something like $~f$, the structure constant of some group rep?
The Kaluza-Klein metric, by reduction, can be written as a ##(4+m) \times (4+m)##symmetric matrix, where ##m## is the dimension of the additional spacetime (if we decompose ##M_D = M_4 \times M_m##). It was show by Bryce de. Witt that, if the non-diagonal metric has the form
$$g_{\mu m} = B^I_{\mu}(x^{\nu}) V^I_m(y^n)$$
And we make a general coordinate transformation
$$\epsilon_m = \lambda^I (x^{\nu}) V_m^I(y^n)$$
We can show that ##\lambda## is the parameter of a non-abelian gauge transformation of ##B_{\mu}##. While every source talks about de Witt derivation, i was not able to find it, indeed, the citations used are not open (B.S. DeWitt, In Relativity, Groups and Topology, eds. C. and B.S. DeWitt (Gordon and Breach, New York, 1964). So i have tried to show it here
$$
\delta g_{rs} = \epsilon^m \partial_m g_{rs} + \partial_{r} \epsilon^m g_{m s} + \partial_s \epsilon^m g_{m s}
$$
$r = \mu, s = n$
$$
\delta g_{\mu n} = \lambda V^{m} \partial_m g_{\mu n} + \partial_{\mu} \lambda V^m g_{mn} + \lambda \partial_n V^m g_{m \mu}
$$
$g_{\mu n} = B_{\mu} V_n$, $g_{mn} = \delta_{mn}$
$$
\delta g_{\mu n} = \delta B_{\mu} V_n + B_{\mu} \delta V_{n} = \lambda B_{\mu} V^m \partial_m V_n + V_n \partial_{\mu} \lambda + \lambda B_{\mu} \partial_n V^m V_m
$$
But ##\delta V_n## is ##0## (honestly, i am just assuming it). Also, i assumed that ##\partial_m V_n V^n = V^n \partial_m V_n##
$$
\delta B_{\mu} V_n = \lambda B_{\mu} V^m \partial_m V_n + V_n \partial_{\mu} \lambda + \lambda B_{\mu} V_m \partial_n V^m
$$
This, in fact, reduces to a "looks like" non-gauge transformation
$$
\delta B_{\mu} = \partial_{\mu} \lambda + (V_n^{-1} V_m \partial_m V_n + V_n^{-1} V_m \partial_n V^m ) \lambda B_{\mu}
$$
But i am not really sure if it is correct. I mean, the term in parenthesis shouldn't be something like $~f$, the structure constant of some group rep?
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