Derivation space as an algebra












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I'm working with the automorphism group that acts on an algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}} < GL({cal A})$. I've seen that the algebra of this group is the space of derivations, that is



$$der({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), y] + [x, a(y)], forall x, y in {cal A}} $$



I can understand that if $a in GL({cal A})$, then $a([x, y]) = a(xy) - a(yx)$ and if $a$ is a derivation, $a(xy) - a(yx) = [a(x), y] + [x, a(y)]$. But how can I prove that this is the algebra of $Aut({cal A})$?



I tried to apply the exponentiation of this algebra and make it act on $[x, y]$ in order to see if the result is the definition of $Aut({cal A})$, but I didn't get it. Any help?






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    I'm working with the automorphism group that acts on an algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}} < GL({cal A})$. I've seen that the algebra of this group is the space of derivations, that is



    $$der({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), y] + [x, a(y)], forall x, y in {cal A}} $$



    I can understand that if $a in GL({cal A})$, then $a([x, y]) = a(xy) - a(yx)$ and if $a$ is a derivation, $a(xy) - a(yx) = [a(x), y] + [x, a(y)]$. But how can I prove that this is the algebra of $Aut({cal A})$?



    I tried to apply the exponentiation of this algebra and make it act on $[x, y]$ in order to see if the result is the definition of $Aut({cal A})$, but I didn't get it. Any help?






    group-theory lie-groups lie-algebras automorphism-group







    share|cite|improve this question



























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      I'm working with the automorphism group that acts on an algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}} < GL({cal A})$. I've seen that the algebra of this group is the space of derivations, that is



      $$der({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), y] + [x, a(y)], forall x, y in {cal A}} $$



      I can understand that if $a in GL({cal A})$, then $a([x, y]) = a(xy) - a(yx)$ and if $a$ is a derivation, $a(xy) - a(yx) = [a(x), y] + [x, a(y)]$. But how can I prove that this is the algebra of $Aut({cal A})$?



      I tried to apply the exponentiation of this algebra and make it act on $[x, y]$ in order to see if the result is the definition of $Aut({cal A})$, but I didn't get it. Any help?






      group-theory lie-groups lie-algebras automorphism-group







      share|cite|improve this question















      I'm working with the automorphism group that acts on an algebra ${cal A}$, $Aut({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), a(y)], forall x, y in {cal A}} < GL({cal A})$. I've seen that the algebra of this group is the space of derivations, that is



      $$der({cal A}) = {a in End({cal A}) | a([x, y]) = [a(x), y] + [x, a(y)], forall x, y in {cal A}} $$



      I can understand that if $a in GL({cal A})$, then $a([x, y]) = a(xy) - a(yx)$ and if $a$ is a derivation, $a(xy) - a(yx) = [a(x), y] + [x, a(y)]$. But how can I prove that this is the algebra of $Aut({cal A})$?



      I tried to apply the exponentiation of this algebra and make it act on $[x, y]$ in order to see if the result is the definition of $Aut({cal A})$, but I didn't get it. Any help?






      group-theory lie-groups lie-algebras automorphism-group




      group-theory lie-groups lie-algebras automorphism-group






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      edited Nov 19 '18 at 0:50

























      asked Nov 19 '18 at 0:44









      Vicky

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          Since the Lie algebra of a group is the tangent space at the identity element, you can write a generic path through the group which goes through the identity, and differentiate. So what I mean is consider a path $g(t)$ in $mathrm{Aut}(mathcal{A})$, say parameterized with $t in (-varepsilon,varepsilon)$, so that $g(0) = mathrm{Id}$.



          The condition that each $g(t)$ is in $mathrm{Aut}(mathcal{A})$ means that for each $t$,



          $$ g(t)[x,y] = [g(t)x,g(t)y] $$



          Let's make up some notation for the sake of doing concrete computations. Let's sake $e_1,dots,e_n$ is a basis of $mathcal{A}$, with structure constants $c^k_{ij}$, so that $[e_i,e_j] = sum_k c^k_{ij} e_k$. Now for our chosen elements $x$ and $y$, let's define functions $a_i(t)$ and $b_j(t)$ so that $g(t)x = sum_i a_i(t)e_i$ and $g(t)y = sum_j b_j(t)e_j$. Then our equation above translates to



          $$ g(t)[x,y] = sum_{i,j,k} c^k_{ij} a_i(t)b_j(t) e_k $$



          Now, let's take the derivative:



          $$ g'(t)[x,y] = sum_{i,j,k} c^k_{ij} left( a_i'(t)b_j(t) + a_i(t)b_j'(t) right) e_k $$



          The right-hand side should translate to the derivation property:



          $$ g'(t)[x,y] = [g'(t)x,y] + [x,g'(t)y] $$






          share|cite|improve this answer





















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            Since the Lie algebra of a group is the tangent space at the identity element, you can write a generic path through the group which goes through the identity, and differentiate. So what I mean is consider a path $g(t)$ in $mathrm{Aut}(mathcal{A})$, say parameterized with $t in (-varepsilon,varepsilon)$, so that $g(0) = mathrm{Id}$.



            The condition that each $g(t)$ is in $mathrm{Aut}(mathcal{A})$ means that for each $t$,



            $$ g(t)[x,y] = [g(t)x,g(t)y] $$



            Let's make up some notation for the sake of doing concrete computations. Let's sake $e_1,dots,e_n$ is a basis of $mathcal{A}$, with structure constants $c^k_{ij}$, so that $[e_i,e_j] = sum_k c^k_{ij} e_k$. Now for our chosen elements $x$ and $y$, let's define functions $a_i(t)$ and $b_j(t)$ so that $g(t)x = sum_i a_i(t)e_i$ and $g(t)y = sum_j b_j(t)e_j$. Then our equation above translates to



            $$ g(t)[x,y] = sum_{i,j,k} c^k_{ij} a_i(t)b_j(t) e_k $$



            Now, let's take the derivative:



            $$ g'(t)[x,y] = sum_{i,j,k} c^k_{ij} left( a_i'(t)b_j(t) + a_i(t)b_j'(t) right) e_k $$



            The right-hand side should translate to the derivation property:



            $$ g'(t)[x,y] = [g'(t)x,y] + [x,g'(t)y] $$






            share|cite|improve this answer


























              1














              Since the Lie algebra of a group is the tangent space at the identity element, you can write a generic path through the group which goes through the identity, and differentiate. So what I mean is consider a path $g(t)$ in $mathrm{Aut}(mathcal{A})$, say parameterized with $t in (-varepsilon,varepsilon)$, so that $g(0) = mathrm{Id}$.



              The condition that each $g(t)$ is in $mathrm{Aut}(mathcal{A})$ means that for each $t$,



              $$ g(t)[x,y] = [g(t)x,g(t)y] $$



              Let's make up some notation for the sake of doing concrete computations. Let's sake $e_1,dots,e_n$ is a basis of $mathcal{A}$, with structure constants $c^k_{ij}$, so that $[e_i,e_j] = sum_k c^k_{ij} e_k$. Now for our chosen elements $x$ and $y$, let's define functions $a_i(t)$ and $b_j(t)$ so that $g(t)x = sum_i a_i(t)e_i$ and $g(t)y = sum_j b_j(t)e_j$. Then our equation above translates to



              $$ g(t)[x,y] = sum_{i,j,k} c^k_{ij} a_i(t)b_j(t) e_k $$



              Now, let's take the derivative:



              $$ g'(t)[x,y] = sum_{i,j,k} c^k_{ij} left( a_i'(t)b_j(t) + a_i(t)b_j'(t) right) e_k $$



              The right-hand side should translate to the derivation property:



              $$ g'(t)[x,y] = [g'(t)x,y] + [x,g'(t)y] $$






              share|cite|improve this answer
























                1












                1








                1






                Since the Lie algebra of a group is the tangent space at the identity element, you can write a generic path through the group which goes through the identity, and differentiate. So what I mean is consider a path $g(t)$ in $mathrm{Aut}(mathcal{A})$, say parameterized with $t in (-varepsilon,varepsilon)$, so that $g(0) = mathrm{Id}$.



                The condition that each $g(t)$ is in $mathrm{Aut}(mathcal{A})$ means that for each $t$,



                $$ g(t)[x,y] = [g(t)x,g(t)y] $$



                Let's make up some notation for the sake of doing concrete computations. Let's sake $e_1,dots,e_n$ is a basis of $mathcal{A}$, with structure constants $c^k_{ij}$, so that $[e_i,e_j] = sum_k c^k_{ij} e_k$. Now for our chosen elements $x$ and $y$, let's define functions $a_i(t)$ and $b_j(t)$ so that $g(t)x = sum_i a_i(t)e_i$ and $g(t)y = sum_j b_j(t)e_j$. Then our equation above translates to



                $$ g(t)[x,y] = sum_{i,j,k} c^k_{ij} a_i(t)b_j(t) e_k $$



                Now, let's take the derivative:



                $$ g'(t)[x,y] = sum_{i,j,k} c^k_{ij} left( a_i'(t)b_j(t) + a_i(t)b_j'(t) right) e_k $$



                The right-hand side should translate to the derivation property:



                $$ g'(t)[x,y] = [g'(t)x,y] + [x,g'(t)y] $$






                share|cite|improve this answer












                Since the Lie algebra of a group is the tangent space at the identity element, you can write a generic path through the group which goes through the identity, and differentiate. So what I mean is consider a path $g(t)$ in $mathrm{Aut}(mathcal{A})$, say parameterized with $t in (-varepsilon,varepsilon)$, so that $g(0) = mathrm{Id}$.



                The condition that each $g(t)$ is in $mathrm{Aut}(mathcal{A})$ means that for each $t$,



                $$ g(t)[x,y] = [g(t)x,g(t)y] $$



                Let's make up some notation for the sake of doing concrete computations. Let's sake $e_1,dots,e_n$ is a basis of $mathcal{A}$, with structure constants $c^k_{ij}$, so that $[e_i,e_j] = sum_k c^k_{ij} e_k$. Now for our chosen elements $x$ and $y$, let's define functions $a_i(t)$ and $b_j(t)$ so that $g(t)x = sum_i a_i(t)e_i$ and $g(t)y = sum_j b_j(t)e_j$. Then our equation above translates to



                $$ g(t)[x,y] = sum_{i,j,k} c^k_{ij} a_i(t)b_j(t) e_k $$



                Now, let's take the derivative:



                $$ g'(t)[x,y] = sum_{i,j,k} c^k_{ij} left( a_i'(t)b_j(t) + a_i(t)b_j'(t) right) e_k $$



                The right-hand side should translate to the derivation property:



                $$ g'(t)[x,y] = [g'(t)x,y] + [x,g'(t)y] $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 19 '18 at 1:07









                Nick

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