What is the interpretation of the Gerstenhaber bracket?

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The homology of an $E_2$-algebra is a Gerstenhaber algebra.



How precisely is the Gerstenhaber structure related to the $E_2$-structure?



Obviously, the Gerstenhaber product is the commutative product that the $E_2$-product induces in homology.



But what precisely is the interpretation of the Gerstenhaber bracket? It cannot be the commutator of the $E_2$-product because that would be zero in homology, right?



Thanks for any hints.




homological-algebra cohomology operads




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    up vote
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    down vote

    favorite
    2












    The homology of an $E_2$-algebra is a Gerstenhaber algebra.



    How precisely is the Gerstenhaber structure related to the $E_2$-structure?



    Obviously, the Gerstenhaber product is the commutative product that the $E_2$-product induces in homology.



    But what precisely is the interpretation of the Gerstenhaber bracket? It cannot be the commutator of the $E_2$-product because that would be zero in homology, right?



    Thanks for any hints.




    homological-algebra cohomology operads




    share|cite|improve this question






















      up vote
      8
      down vote

      favorite
      2









      up vote
      8
      down vote

      favorite
      2






      2





      The homology of an $E_2$-algebra is a Gerstenhaber algebra.



      How precisely is the Gerstenhaber structure related to the $E_2$-structure?



      Obviously, the Gerstenhaber product is the commutative product that the $E_2$-product induces in homology.



      But what precisely is the interpretation of the Gerstenhaber bracket? It cannot be the commutator of the $E_2$-product because that would be zero in homology, right?



      Thanks for any hints.




      homological-algebra cohomology operads




      share|cite|improve this question












      The homology of an $E_2$-algebra is a Gerstenhaber algebra.



      How precisely is the Gerstenhaber structure related to the $E_2$-structure?



      Obviously, the Gerstenhaber product is the commutative product that the $E_2$-product induces in homology.



      But what precisely is the interpretation of the Gerstenhaber bracket? It cannot be the commutator of the $E_2$-product because that would be zero in homology, right?



      Thanks for any hints.





      homological-algebra cohomology operads





      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 12 at 18:50









      Lukas Woike

      5036




      5036




















          1 Answer
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          I think the most transparent interpretation is by identifying $E_2$ algebras with brace algebras (which was proved by McClure and Smith).



          Namely, a brace algebra satisfies the relation
          $$ab - (-1)^aba = (-1)^ d(ab) - (-1)^(da)b + adb,$$
          where $ab$ is one of the brace operations (so, it witnesses the first-order commutativity of the multiplication). The Gerstenhaber bracket is the antisymmetrization of the first brace:
          $$[a, b] = ab - (-1)^a ba.$$
          Note that from the first equation you can see that $[-, -]$ is a $d$-closed operation, so it makes sense on homology.



          So, the Gerstenhaber bracket is a secondary operation: the product is commutative on homology, but the homotopy might not be (anti)symmetric.






          share|cite|improve this answer




















          • Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket?
            – Lukas Woike
            Aug 13 at 7:46










          • Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)?
            – Lukas Woike
            Aug 13 at 7:46






          • 1




            For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $ab$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $ba$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie1rightarrow H_bullet(E_2)$ in arity 2 sends the bracket to the fundamental class.
            – Pavel Safronov
            Aug 13 at 8:15










          • May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_p,q : A_p otimes A_q to A_p+q+1$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$?
            – Lukas Woike
            Aug 15 at 7:57










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          1 Answer
          1






          active

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          active

          oldest

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          up vote
          12
          down vote



          accepted










          I think the most transparent interpretation is by identifying $E_2$ algebras with brace algebras (which was proved by McClure and Smith).



          Namely, a brace algebra satisfies the relation
          $$ab - (-1)^aba = (-1)^ d(ab) - (-1)^(da)b + adb,$$
          where $ab$ is one of the brace operations (so, it witnesses the first-order commutativity of the multiplication). The Gerstenhaber bracket is the antisymmetrization of the first brace:
          $$[a, b] = ab - (-1)^a ba.$$
          Note that from the first equation you can see that $[-, -]$ is a $d$-closed operation, so it makes sense on homology.



          So, the Gerstenhaber bracket is a secondary operation: the product is commutative on homology, but the homotopy might not be (anti)symmetric.






          share|cite|improve this answer




















          • Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket?
            – Lukas Woike
            Aug 13 at 7:46










          • Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)?
            – Lukas Woike
            Aug 13 at 7:46






          • 1




            For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $ab$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $ba$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie1rightarrow H_bullet(E_2)$ in arity 2 sends the bracket to the fundamental class.
            – Pavel Safronov
            Aug 13 at 8:15










          • May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_p,q : A_p otimes A_q to A_p+q+1$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$?
            – Lukas Woike
            Aug 15 at 7:57














          up vote
          12
          down vote



          accepted










          I think the most transparent interpretation is by identifying $E_2$ algebras with brace algebras (which was proved by McClure and Smith).



          Namely, a brace algebra satisfies the relation
          $$ab - (-1)^aba = (-1)^ d(ab) - (-1)^(da)b + adb,$$
          where $ab$ is one of the brace operations (so, it witnesses the first-order commutativity of the multiplication). The Gerstenhaber bracket is the antisymmetrization of the first brace:
          $$[a, b] = ab - (-1)^a ba.$$
          Note that from the first equation you can see that $[-, -]$ is a $d$-closed operation, so it makes sense on homology.



          So, the Gerstenhaber bracket is a secondary operation: the product is commutative on homology, but the homotopy might not be (anti)symmetric.






          share|cite|improve this answer




















          • Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket?
            – Lukas Woike
            Aug 13 at 7:46










          • Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)?
            – Lukas Woike
            Aug 13 at 7:46






          • 1




            For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $ab$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $ba$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie1rightarrow H_bullet(E_2)$ in arity 2 sends the bracket to the fundamental class.
            – Pavel Safronov
            Aug 13 at 8:15










          • May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_p,q : A_p otimes A_q to A_p+q+1$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$?
            – Lukas Woike
            Aug 15 at 7:57












          up vote
          12
          down vote



          accepted







          up vote
          12
          down vote



          accepted






          I think the most transparent interpretation is by identifying $E_2$ algebras with brace algebras (which was proved by McClure and Smith).



          Namely, a brace algebra satisfies the relation
          $$ab - (-1)^aba = (-1)^ d(ab) - (-1)^(da)b + adb,$$
          where $ab$ is one of the brace operations (so, it witnesses the first-order commutativity of the multiplication). The Gerstenhaber bracket is the antisymmetrization of the first brace:
          $$[a, b] = ab - (-1)^a ba.$$
          Note that from the first equation you can see that $[-, -]$ is a $d$-closed operation, so it makes sense on homology.



          So, the Gerstenhaber bracket is a secondary operation: the product is commutative on homology, but the homotopy might not be (anti)symmetric.






          share|cite|improve this answer












          I think the most transparent interpretation is by identifying $E_2$ algebras with brace algebras (which was proved by McClure and Smith).



          Namely, a brace algebra satisfies the relation
          $$ab - (-1)^aba = (-1)^ d(ab) - (-1)^(da)b + adb,$$
          where $ab$ is one of the brace operations (so, it witnesses the first-order commutativity of the multiplication). The Gerstenhaber bracket is the antisymmetrization of the first brace:
          $$[a, b] = ab - (-1)^a ba.$$
          Note that from the first equation you can see that $[-, -]$ is a $d$-closed operation, so it makes sense on homology.



          So, the Gerstenhaber bracket is a secondary operation: the product is commutative on homology, but the homotopy might not be (anti)symmetric.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 12 at 19:22









          Pavel Safronov

          2,41711013




          2,41711013











          • Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket?
            – Lukas Woike
            Aug 13 at 7:46










          • Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)?
            – Lukas Woike
            Aug 13 at 7:46






          • 1




            For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $ab$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $ba$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie1rightarrow H_bullet(E_2)$ in arity 2 sends the bracket to the fundamental class.
            – Pavel Safronov
            Aug 13 at 8:15










          • May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_p,q : A_p otimes A_q to A_p+q+1$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$?
            – Lukas Woike
            Aug 15 at 7:57
















          • Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket?
            – Lukas Woike
            Aug 13 at 7:46










          • Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)?
            – Lukas Woike
            Aug 13 at 7:46






          • 1




            For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $ab$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $ba$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie1rightarrow H_bullet(E_2)$ in arity 2 sends the bracket to the fundamental class.
            – Pavel Safronov
            Aug 13 at 8:15










          • May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_p,q : A_p otimes A_q to A_p+q+1$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$?
            – Lukas Woike
            Aug 15 at 7:57















          Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket?
          – Lukas Woike
          Aug 13 at 7:46




          Thank you for your answer. So if I have the homotopy which makes the multiplication commutative, can I construct the Gerstenhaber bracket?
          – Lukas Woike
          Aug 13 at 7:46












          Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)?
          – Lukas Woike
          Aug 13 at 7:46




          Is there also a possibility to relate the Gerstenhaber bracket to "swapping factors twice" (to some kind of double braiding, using the terminology of $E_2$-categories)?
          – Lukas Woike
          Aug 13 at 7:46




          1




          1




          For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $ab$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $ba$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie1rightarrow H_bullet(E_2)$ in arity 2 sends the bracket to the fundamental class.
          – Pavel Safronov
          Aug 13 at 8:15




          For the first question: yes, if the homotopy satisfies the pre-Lie relation (possibly up to homotopy). For the second question: I guess that's indeed how you can think about it. If you imagine multiplications in $E_2$ parametrized by $S^1$, then the homotopy $ab$ is the class of a semicircle from the basepoint to its antipodal point. The homotopy $ba$ is the class of the other semicircle. Adding them together you get the fundamental class of $S^1$. Here I am using that $Lie1rightarrow H_bullet(E_2)$ in arity 2 sends the bracket to the fundamental class.
          – Pavel Safronov
          Aug 13 at 8:15












          May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_p,q : A_p otimes A_q to A_p+q+1$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$?
          – Lukas Woike
          Aug 15 at 7:57




          May I ask about more details: My $E_2$-algebra $A$ is a chain complex (not cochain), and I have a homotopy between multiplication and opposite multiplication; it consists of maps $h_p,q : A_p otimes A_q to A_p+q+1$. Now what would your pre-Lie relation be and how can I build the braces or $[-,-]$ from $h$?
          – Lukas Woike
          Aug 15 at 7:57

















           

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