Involution action on $H^3(S^1times S^2)$












1












$begingroup$


I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^3(S^1times S^2)=mathbb{R}.$$



I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
Involution action on $H^1(S^1times S^2)$.




However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




Any help would be appreciated! Thank you in advance!






group-actions de-rham-cohomology involutions







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^3(S^1times S^2)=mathbb{R}.$$



    I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
    Involution action on $H^1(S^1times S^2)$.




    However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



    I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




    Any help would be appreciated! Thank you in advance!






    group-actions de-rham-cohomology involutions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^3(S^1times S^2)=mathbb{R}.$$



      I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
      Involution action on $H^1(S^1times S^2)$.




      However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



      I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




      Any help would be appreciated! Thank you in advance!






      group-actions de-rham-cohomology involutions







      share|cite|improve this question









      $endgroup$




      I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^3(S^1times S^2)=mathbb{R}.$$



      I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
      Involution action on $H^1(S^1times S^2)$.




      However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



      I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




      Any help would be appreciated! Thank you in advance!






      group-actions de-rham-cohomology involutions




      group-actions de-rham-cohomology involutions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 24 '18 at 6:11









      Lev BanLev Ban

      1,0721317




      1,0721317






















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19













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          2












          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19


















          2












          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19
















          2












          2








          2





          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$



          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 22:02









          Michael AlbaneseMichael Albanese

          63.8k1599309




          63.8k1599309








          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19
















          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19










          1




          1




          $begingroup$
          More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
          $endgroup$
          – Pedro Tamaroff♦
          Dec 24 '18 at 22:18




          $begingroup$
          More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
          $endgroup$
          – Pedro Tamaroff♦
          Dec 24 '18 at 22:18












          $begingroup$
          @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 19:48




          $begingroup$
          @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 19:48












          $begingroup$
          @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
          $endgroup$
          – Michael Albanese
          Dec 26 '18 at 21:29




          $begingroup$
          @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
          $endgroup$
          – Michael Albanese
          Dec 26 '18 at 21:29












          $begingroup$
          @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 23:38






          $begingroup$
          @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 23:38






          1




          1




          $begingroup$
          The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
          $endgroup$
          – Michael Albanese
          Dec 27 '18 at 1:19






          $begingroup$
          The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
          $endgroup$
          – Michael Albanese
          Dec 27 '18 at 1:19




















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          Wiesbaden

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          Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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          Marschland

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          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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          Dieringhausen

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          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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