<p>In this work, we introduce a method for finding exact solutions to the vacuum Einstein field equations in higher dimensions from a given solution to the chiral equation. When considering a <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq4.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="40" /> </InlineMediaObject> <EquationSource Format="TEX">\(n + 2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>+</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>-dimensional spacetime with <i>n</i> commutative Killing vectors, the metric tensor can take the form <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq5.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="291" /> </InlineMediaObject> <EquationSource Format="TEX">\(\hat{g} = f ( \rho , \zeta ) ( d \rho ^2 + d \zeta ^2 ) + g_{\mu \nu } ( \rho , \zeta ) d x^\mu d x^\nu \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mover accent="true"> <mi>g</mi> <mo stretchy="false">^</mo> </mover> <mo>=</mo> <mi>f</mi> <mrow> <mo stretchy="false">(</mo> <mi>ρ</mi> <mo>,</mo> <mi>ζ</mi> <mo stretchy="false">)</mo> </mrow> <mrow> <mo stretchy="false">(</mo> <mi>d</mi> <msup> <mi>ρ</mi> <mn>2</mn> </msup> <mo>+</mo> <mi>d</mi> <msup> <mi>ζ</mi> <mn>2</mn> </msup> <mo stretchy="false">)</mo> </mrow> <mo>+</mo> <msub> <mi>g</mi> <mrow> <mi>μ</mi> <mi>ν</mi> </mrow> </msub> <mrow> <mo stretchy="false">(</mo> <mi>ρ</mi> <mo>,</mo> <mi>ζ</mi> <mo stretchy="false">)</mo> </mrow> <mi>d</mi> <msup> <mi>x</mi> <mi>μ</mi> </msup> <mi>d</mi> <msup> <mi>x</mi> <mi>ν</mi> </msup> </mrow> </math></EquationSource> </InlineEquation>. Then, the Einstein field equations in vacuum reduce to a chiral equation, <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq6.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="197" /> </InlineMediaObject> <EquationSource Format="TEX">\(( \rho g_{, z} g ^{-1} )_{, \bar{z}} + ( \rho g_{, \bar{z}} g ^{-1} )_{, z} = 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mrow> <mo stretchy="false">(</mo> <mi>ρ</mi> <msub> <mi>g</mi> <mrow> <mo>,</mo> <mi>z</mi> </mrow> </msub> <msup> <mi>g</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo stretchy="false">)</mo> </mrow> <mrow> <mo>,</mo> <mover accent="true"> <mrow> <mi>z</mi> </mrow> <mrow> <mo stretchy="false">¯</mo> </mrow> </mover> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mo stretchy="false">(</mo> <mi>ρ</mi> <msub> <mi>g</mi> <mrow> <mo>,</mo> <mover accent="true"> <mrow> <mi>z</mi> </mrow> <mrow> <mo stretchy="false">¯</mo> </mrow> </mover> </mrow> </msub> <msup> <mi>g</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo stretchy="false">)</mo> </mrow> <mrow> <mo>,</mo> <mi>z</mi> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, and two differential equations, <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq7.gif" Format="GIF" Height="23" Rendition="HTML" Resolution="72" Type="Linedraw" Width="203" /> </InlineMediaObject> <EquationSource Format="TEX">\(( \ln f \rho ^{1-1/n} )_{, Z} = \frac{\rho }{2} \operatorname {tr} ( g_{, _Z} g^{-1} )^2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mrow> <mo stretchy="false">(</mo> <mo>ln</mo> <mi>f</mi> <msup> <mi>ρ</mi> <mrow> <mn>1</mn> <mo>-</mo> <mn>1</mn> <mo stretchy="false">/</mo> <mi>n</mi> </mrow> </msup> <mo stretchy="false">)</mo> </mrow> <mrow> <mo>,</mo> <mi>Z</mi> </mrow> </msub> <mo>=</mo> <mfrac> <mi>ρ</mi> <mn>2</mn> </mfrac> <mo>tr</mo> <msup> <mrow> <mo stretchy="false">(</mo> <msub> <mi>g</mi> <msub> <mo>,</mo> <mi>Z</mi> </msub> </msub> <msup> <mi>g</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo stretchy="false">)</mo> </mrow> <mn>2</mn> </msup> </mrow> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq8.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="95" /> </InlineMediaObject> <EquationSource Format="TEX">\(g \in SL( n, \mathbb {R} )\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>g</mi> <mo>∈</mo> <mi>S</mi> <mi>L</mi> <mo stretchy="false">(</mo> <mi>n</mi> <mo>,</mo> <mi mathvariant="double-struck">R</mi> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> is the normalized matrix representation of <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq9.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="25" /> </InlineMediaObject> <EquationSource Format="TEX">\(g_{\mu \nu }\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>g</mi> <mrow> <mi>μ</mi> <mi>ν</mi> </mrow> </msub> </math></EquationSource> </InlineEquation>, <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq10.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="77" /> </InlineMediaObject> <EquationSource Format="TEX">\(z = \rho + i \zeta \)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>z</mi> <mo>=</mo> <mi>ρ</mi> <mo>+</mo> <mi>i</mi> <mi>ζ</mi> </mrow> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq11.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="62" /> </InlineMediaObject> <EquationSource Format="TEX">\(Z = z, \bar{z}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>Z</mi> <mo>=</mo> <mi>z</mi> <mo>,</mo> <mover accent="true"> <mrow> <mi>z</mi> </mrow> <mrow> <mo stretchy="false">¯</mo> </mrow> </mover> </mrow> </math></EquationSource> </InlineEquation>. We use the ansatz <InlineEquation ID="IEq12"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq12.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="69" /> </InlineMediaObject> <EquationSource Format="TEX">\(g = g ( \xi ^a )\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>g</mi> <mo>=</mo> <mi>g</mi> <mo stretchy="false">(</mo> <msup> <mi>ξ</mi> <mi>a</mi> </msup> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>, where the parameters <InlineEquation ID="IEq13"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq13.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="18" /> </InlineMediaObject> <EquationSource Format="TEX">\(\xi ^a\)</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>ξ</mi> <mi>a</mi> </msup> </math></EquationSource> </InlineEquation> depend on <i>z</i> and <InlineEquation ID="IEq14"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq14.gif" Format="GIF" Height="12" Rendition="HTML" Resolution="72" Type="Linedraw" Width="11" /> </InlineMediaObject> <EquationSource Format="TEX">\(\bar{z}\)</EquationSource> <EquationSource Format="MATHML"><math> <mover accent="true"> <mrow> <mi>z</mi> </mrow> <mrow> <mo stretchy="false">¯</mo> </mrow> </mover> </math></EquationSource> </InlineEquation> and satisfy a generalized Laplace equation, <InlineEquation ID="IEq15"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq15.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="147" /> </InlineMediaObject> <EquationSource Format="TEX">\(( \rho \xi ^a _{, z} )_{, \bar{z}} + ( \rho \xi ^a _{, \bar{z}} )_{, z} = 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mrow> <mo stretchy="false">(</mo> <mi>ρ</mi> <msubsup> <mi>ξ</mi> <mrow> <mo>,</mo> <mi>z</mi> </mrow> <mi>a</mi> </msubsup> <mo stretchy="false">)</mo> </mrow> <mrow> <mo>,</mo> <mover accent="true"> <mrow> <mi>z</mi> </mrow> <mrow> <mo stretchy="false">¯</mo> </mrow> </mover> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mo stretchy="false">(</mo> <mi>ρ</mi> <msubsup> <mi>ξ</mi> <mrow> <mo>,</mo> <mover accent="true"> <mrow> <mi>z</mi> </mrow> <mrow> <mo stretchy="false">¯</mo> </mrow> </mover> </mrow> <mi>a</mi> </msubsup> <mo stretchy="false">)</mo> </mrow> <mrow> <mo>,</mo> <mi>z</mi> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>. The chiral equation to the Killing equation, <InlineEquation ID="IEq16"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq16.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="117" /> </InlineMediaObject> <EquationSource Format="TEX">\(A_{a, \xi ^b} + A_{b, \xi ^a} = 0\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>A</mi> <mrow> <mi>a</mi> <mo>,</mo> <msup> <mi>ξ</mi> <mi>b</mi> </msup> </mrow> </msub> <mo>+</mo> <msub> <mi>A</mi> <mrow> <mi>b</mi> <mo>,</mo> <msup> <mi>ξ</mi> <mi>a</mi> </msup> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </math></EquationSource> </InlineEquation>, where <InlineEquation ID="IEq17"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq17.gif" Format="GIF" Height="21" Rendition="HTML" Resolution="72" Type="Linedraw" Width="90" /> </InlineMediaObject> <EquationSource Format="TEX">\(A_a = g_{, \xi ^a} g^{-1}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <msub> <mi>A</mi> <mi>a</mi> </msub> <mo>=</mo> <msub> <mi>g</mi> <mrow> <mo>,</mo> <msup> <mi>ξ</mi> <mi>a</mi> </msup> </mrow> </msub> <msup> <mi>g</mi> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </math></EquationSource> </InlineEquation>. Furthermore, we assume that the matrices <InlineEquation ID="IEq18"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10714_2025_3467_Article_IEq18.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="22" /> </InlineMediaObject> <EquationSource Format="TEX">\(A_a\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>A</mi> <mi>a</mi> </msub> </math></EquationSource> </InlineEquation> commute with each other; in this way, they fulfill the Killing equation.</p>

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Flat subspaces of the \(SL(n,\mathbb {R})\) chiral equations

  • I. A. Sarmiento-Alvarado,
  • P. Wiederhold,
  • T. Matos

摘要

In this work, we introduce a method for finding exact solutions to the vacuum Einstein field equations in higher dimensions from a given solution to the chiral equation. When considering a \(n + 2\) n + 2 -dimensional spacetime with n commutative Killing vectors, the metric tensor can take the form \(\hat{g} = f ( \rho , \zeta ) ( d \rho ^2 + d \zeta ^2 ) + g_{\mu \nu } ( \rho , \zeta ) d x^\mu d x^\nu \) g ^ = f ( ρ , ζ ) ( d ρ 2 + d ζ 2 ) + g μ ν ( ρ , ζ ) d x μ d x ν . Then, the Einstein field equations in vacuum reduce to a chiral equation, \(( \rho g_{, z} g ^{-1} )_{, \bar{z}} + ( \rho g_{, \bar{z}} g ^{-1} )_{, z} = 0\) ( ρ g , z g - 1 ) , z ¯ + ( ρ g , z ¯ g - 1 ) , z = 0 , and two differential equations, \(( \ln f \rho ^{1-1/n} )_{, Z} = \frac{\rho }{2} \operatorname {tr} ( g_{, _Z} g^{-1} )^2\) ( ln f ρ 1 - 1 / n ) , Z = ρ 2 tr ( g , Z g - 1 ) 2 , where \(g \in SL( n, \mathbb {R} )\) g S L ( n , R ) is the normalized matrix representation of \(g_{\mu \nu }\) g μ ν , \(z = \rho + i \zeta \) z = ρ + i ζ and \(Z = z, \bar{z}\) Z = z , z ¯ . We use the ansatz \(g = g ( \xi ^a )\) g = g ( ξ a ) , where the parameters \(\xi ^a\) ξ a depend on z and \(\bar{z}\) z ¯ and satisfy a generalized Laplace equation, \(( \rho \xi ^a _{, z} )_{, \bar{z}} + ( \rho \xi ^a _{, \bar{z}} )_{, z} = 0\) ( ρ ξ , z a ) , z ¯ + ( ρ ξ , z ¯ a ) , z = 0 . The chiral equation to the Killing equation, \(A_{a, \xi ^b} + A_{b, \xi ^a} = 0\) A a , ξ b + A b , ξ a = 0 , where \(A_a = g_{, \xi ^a} g^{-1}\) A a = g , ξ a g - 1 . Furthermore, we assume that the matrices \(A_a\) A a commute with each other; in this way, they fulfill the Killing equation.