Mixing
About involution
$begingroup$
Below is an excerpt from chapter 7 of Winfried Scharlau's Quadratic and Hermitian Forms:
I understand that how to convert left module to right module by involution.But what is meaning of we always interpret dual module M* as right module in this sense? Also what is meaning of scalar multiplication is twisted by involution?
linear-algebra abstract-algebra algebraic-geometry
edited Jan 23 at 8:46
user1551
73.5k566129
asked Jan 23 at 7:51
maths studentmaths student
6251521
$endgroup$
add a comment |
$begingroup$
Below is an excerpt from chapter 7 of Winfried Scharlau's Quadratic and Hermitian Forms:
I understand that how to convert left module to right module by involution.But what is meaning of we always interpret dual module M* as right module in this sense? Also what is meaning of scalar multiplication is twisted by involution?
linear-algebra abstract-algebra algebraic-geometry
edited Jan 23 at 8:46
user1551
73.5k566129
asked Jan 23 at 7:51
maths studentmaths student
6251521
$endgroup$
add a comment |
$begingroup$
Below is an excerpt from chapter 7 of Winfried Scharlau's Quadratic and Hermitian Forms:
I understand that how to convert left module to right module by involution.But what is meaning of we always interpret dual module M* as right module in this sense? Also what is meaning of scalar multiplication is twisted by involution?
linear-algebra abstract-algebra algebraic-geometry
edited Jan 23 at 8:46
user1551
73.5k566129
asked Jan 23 at 7:51
maths studentmaths student
6251521
$endgroup$
Below is an excerpt from chapter 7 of Winfried Scharlau's Quadratic and Hermitian Forms:
I understand that how to convert left module to right module by involution.But what is meaning of we always interpret dual module M* as right module in this sense? Also what is meaning of scalar multiplication is twisted by involution?
linear-algebra abstract-algebra algebraic-geometry
linear-algebra abstract-algebra algebraic-geometry
edited Jan 23 at 8:46
user1551
73.5k566129
asked Jan 23 at 7:51
maths studentmaths student
6251521
edited Jan 23 at 8:46
user1551
73.5k566129
asked Jan 23 at 7:51
maths studentmaths student
6251521
edited Jan 23 at 8:46
user1551
73.5k566129
edited Jan 23 at 8:46
user1551
73.5k566129
edited Jan 23 at 8:46
user1551
73.5k566129
73.5k566129
asked Jan 23 at 7:51
maths studentmaths student
6251521
asked Jan 23 at 7:51
maths studentmaths student
6251521
asked Jan 23 at 7:51
maths studentmaths student
6251521
6251521
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When we work over a commutative ring $R$ and $M$ is a left $R$-module, the dual module $M^*$ is considered to be Hom$(M,R)$ with right $R$-module structure.
If we now take $R$ to be not necessarily commutative, then $M^*=$Hom$(M,R)$ has, a priori, only a left $R$-module structure, given by $(alpha f)x=alpha(fx)$ for $fin M^*$, $xin M$, $alphain R$. We'd want to put a right $R$-module structure on $M^*$, and we can do when $R$ has an involution $*$, for then we can "twist" the left action to a right one, with the cost of changing the scalar to its conjugate via the involution: we define $$fcdotalpha:=alpha^* f,$$
so that $(falpha)(x)=alpha^*(fx)$ for $fin M^*$, $xin M$, $alphain R$.
Now suppose that $R$ is commutative with a nontrivial involution and $M$ is a left $R$-module. We have two ways of taking the dual of $M$: one is by considering the natural right action of $R$ on $M$, which would give $falpha=alpha f$; the other is by considering the right action coming from twisting the left action via the involution, which would give $falpha=alpha^* f$. You can see that they are different (e.g. if $R=mathbb{C}$ and $alpha=i$ then the second action has a negative sign with respect to the first). So we have to be careful in this case, because we could be giving the same name to two different objects.
Edited to address further questions
1) We have $(Moplus N)^*=M^*oplus N^*$: by definition, if $fin(Moplus N)^*$ then $f:Moplus Nrightarrow R$. Consider the projections $f_M:Mrightarrow R$, $f_N:Nrightarrow R$ such that $f(m+n)=f_M(m)+f_N(n)$ for $min M, nin N$, and the right $R$-module structures are compatible. This shows that $(Moplus N)^*subseteq M^*oplus N^*$. Conversely, if $f_Min M^*$ and $f_Nin N^*$ then $f:=f_M+f_Nin (Moplus N)^*$ and all operations go well, so the two modules are the same.
2) Let us see that $Rcong R^*$ as right modules: We suppose that $R$ has identity. If the action was not twisted, we could use the regular representation $phi:Rrightarrow$Hom$(R,R)$, $phi(r)=phi_r$ such that $phi_r(x):=xr$, but due to the twisting caused by the involution we are going to consider $phi_r(x):=r^*x$.
First we see that $phi$ is a homomorphism of right $R$-modules. If $r,sin R$ then $phi_{r+s}(x)=(r+s)^*x=(r^*+s^*)x=r^*x+s^*x=phi_r(x)+phi_s(x)$ for all $xin R$ implies that $phi(r+s)=phi(r)+phi(s)$, and $phi_{sr}(x)=(sr)^*(x)=(r^*s^*)x=r^*(s^*x)=r^*phi_s(x)=phi_s(x)cdot r$ implies that $phi(sr)=phi(s)cdot r$.
Now wee see that $phi$ is injective: since it is additive it is enough tho show that $phi(r)=0$ implies $r=0$. This is true because $phi_r(x)=0$ for all $x$ implies $r^*x=0$ for all $x$, in particular $r^*=r^*1=0$, hence $r=0$.
Finally, we show that $phi$ is surjective: Given $fin$Hom$(R,R)$ note that $f(x)=f(1x)=f(1)x$, so $f$ is determined by $f(1)$; hence $phi(f(1)^*)=f$ since $phi_{f(1)^*}(x)=(f(1)^*)^*x=f(1)x=f(x)$ for all $xin R$.
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
@ninjahatori Please, see the update in my answer.
$endgroup$
– Jose Brox
Jan 23 at 10:46
$begingroup$
@ninjahatori OK, let's see if I can help you!
$endgroup$
– Jose Brox
Jan 28 at 11:23
add a comment |
Your Answer
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1 Answer
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$begingroup$
When we work over a commutative ring $R$ and $M$ is a left $R$-module, the dual module $M^*$ is considered to be Hom$(M,R)$ with right $R$-module structure.
If we now take $R$ to be not necessarily commutative, then $M^*=$Hom$(M,R)$ has, a priori, only a left $R$-module structure, given by $(alpha f)x=alpha(fx)$ for $fin M^*$, $xin M$, $alphain R$. We'd want to put a right $R$-module structure on $M^*$, and we can do when $R$ has an involution $*$, for then we can "twist" the left action to a right one, with the cost of changing the scalar to its conjugate via the involution: we define $$fcdotalpha:=alpha^* f,$$
so that $(falpha)(x)=alpha^*(fx)$ for $fin M^*$, $xin M$, $alphain R$.
Now suppose that $R$ is commutative with a nontrivial involution and $M$ is a left $R$-module. We have two ways of taking the dual of $M$: one is by considering the natural right action of $R$ on $M$, which would give $falpha=alpha f$; the other is by considering the right action coming from twisting the left action via the involution, which would give $falpha=alpha^* f$. You can see that they are different (e.g. if $R=mathbb{C}$ and $alpha=i$ then the second action has a negative sign with respect to the first). So we have to be careful in this case, because we could be giving the same name to two different objects.
Edited to address further questions
1) We have $(Moplus N)^*=M^*oplus N^*$: by definition, if $fin(Moplus N)^*$ then $f:Moplus Nrightarrow R$. Consider the projections $f_M:Mrightarrow R$, $f_N:Nrightarrow R$ such that $f(m+n)=f_M(m)+f_N(n)$ for $min M, nin N$, and the right $R$-module structures are compatible. This shows that $(Moplus N)^*subseteq M^*oplus N^*$. Conversely, if $f_Min M^*$ and $f_Nin N^*$ then $f:=f_M+f_Nin (Moplus N)^*$ and all operations go well, so the two modules are the same.
2) Let us see that $Rcong R^*$ as right modules: We suppose that $R$ has identity. If the action was not twisted, we could use the regular representation $phi:Rrightarrow$Hom$(R,R)$, $phi(r)=phi_r$ such that $phi_r(x):=xr$, but due to the twisting caused by the involution we are going to consider $phi_r(x):=r^*x$.
First we see that $phi$ is a homomorphism of right $R$-modules. If $r,sin R$ then $phi_{r+s}(x)=(r+s)^*x=(r^*+s^*)x=r^*x+s^*x=phi_r(x)+phi_s(x)$ for all $xin R$ implies that $phi(r+s)=phi(r)+phi(s)$, and $phi_{sr}(x)=(sr)^*(x)=(r^*s^*)x=r^*(s^*x)=r^*phi_s(x)=phi_s(x)cdot r$ implies that $phi(sr)=phi(s)cdot r$.
Now wee see that $phi$ is injective: since it is additive it is enough tho show that $phi(r)=0$ implies $r=0$. This is true because $phi_r(x)=0$ for all $x$ implies $r^*x=0$ for all $x$, in particular $r^*=r^*1=0$, hence $r=0$.
Finally, we show that $phi$ is surjective: Given $fin$Hom$(R,R)$ note that $f(x)=f(1x)=f(1)x$, so $f$ is determined by $f(1)$; hence $phi(f(1)^*)=f$ since $phi_{f(1)^*}(x)=(f(1)^*)^*x=f(1)x=f(x)$ for all $xin R$.
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
@ninjahatori Please, see the update in my answer.
$endgroup$
– Jose Brox
Jan 23 at 10:46
$begingroup$
@ninjahatori OK, let's see if I can help you!
$endgroup$
– Jose Brox
Jan 28 at 11:23
add a comment |
$begingroup$
When we work over a commutative ring $R$ and $M$ is a left $R$-module, the dual module $M^*$ is considered to be Hom$(M,R)$ with right $R$-module structure.
If we now take $R$ to be not necessarily commutative, then $M^*=$Hom$(M,R)$ has, a priori, only a left $R$-module structure, given by $(alpha f)x=alpha(fx)$ for $fin M^*$, $xin M$, $alphain R$. We'd want to put a right $R$-module structure on $M^*$, and we can do when $R$ has an involution $*$, for then we can "twist" the left action to a right one, with the cost of changing the scalar to its conjugate via the involution: we define $$fcdotalpha:=alpha^* f,$$
so that $(falpha)(x)=alpha^*(fx)$ for $fin M^*$, $xin M$, $alphain R$.
Now suppose that $R$ is commutative with a nontrivial involution and $M$ is a left $R$-module. We have two ways of taking the dual of $M$: one is by considering the natural right action of $R$ on $M$, which would give $falpha=alpha f$; the other is by considering the right action coming from twisting the left action via the involution, which would give $falpha=alpha^* f$. You can see that they are different (e.g. if $R=mathbb{C}$ and $alpha=i$ then the second action has a negative sign with respect to the first). So we have to be careful in this case, because we could be giving the same name to two different objects.
Edited to address further questions
1) We have $(Moplus N)^*=M^*oplus N^*$: by definition, if $fin(Moplus N)^*$ then $f:Moplus Nrightarrow R$. Consider the projections $f_M:Mrightarrow R$, $f_N:Nrightarrow R$ such that $f(m+n)=f_M(m)+f_N(n)$ for $min M, nin N$, and the right $R$-module structures are compatible. This shows that $(Moplus N)^*subseteq M^*oplus N^*$. Conversely, if $f_Min M^*$ and $f_Nin N^*$ then $f:=f_M+f_Nin (Moplus N)^*$ and all operations go well, so the two modules are the same.
2) Let us see that $Rcong R^*$ as right modules: We suppose that $R$ has identity. If the action was not twisted, we could use the regular representation $phi:Rrightarrow$Hom$(R,R)$, $phi(r)=phi_r$ such that $phi_r(x):=xr$, but due to the twisting caused by the involution we are going to consider $phi_r(x):=r^*x$.
First we see that $phi$ is a homomorphism of right $R$-modules. If $r,sin R$ then $phi_{r+s}(x)=(r+s)^*x=(r^*+s^*)x=r^*x+s^*x=phi_r(x)+phi_s(x)$ for all $xin R$ implies that $phi(r+s)=phi(r)+phi(s)$, and $phi_{sr}(x)=(sr)^*(x)=(r^*s^*)x=r^*(s^*x)=r^*phi_s(x)=phi_s(x)cdot r$ implies that $phi(sr)=phi(s)cdot r$.
Now wee see that $phi$ is injective: since it is additive it is enough tho show that $phi(r)=0$ implies $r=0$. This is true because $phi_r(x)=0$ for all $x$ implies $r^*x=0$ for all $x$, in particular $r^*=r^*1=0$, hence $r=0$.
Finally, we show that $phi$ is surjective: Given $fin$Hom$(R,R)$ note that $f(x)=f(1x)=f(1)x$, so $f$ is determined by $f(1)$; hence $phi(f(1)^*)=f$ since $phi_{f(1)^*}(x)=(f(1)^*)^*x=f(1)x=f(x)$ for all $xin R$.
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
$endgroup$
$begingroup$
@ninjahatori Please, see the update in my answer.
$endgroup$
– Jose Brox
Jan 23 at 10:46
$begingroup$
@ninjahatori OK, let's see if I can help you!
$endgroup$
– Jose Brox
Jan 28 at 11:23
add a comment |
$begingroup$
When we work over a commutative ring $R$ and $M$ is a left $R$-module, the dual module $M^*$ is considered to be Hom$(M,R)$ with right $R$-module structure.
If we now take $R$ to be not necessarily commutative, then $M^*=$Hom$(M,R)$ has, a priori, only a left $R$-module structure, given by $(alpha f)x=alpha(fx)$ for $fin M^*$, $xin M$, $alphain R$. We'd want to put a right $R$-module structure on $M^*$, and we can do when $R$ has an involution $*$, for then we can "twist" the left action to a right one, with the cost of changing the scalar to its conjugate via the involution: we define $$fcdotalpha:=alpha^* f,$$
so that $(falpha)(x)=alpha^*(fx)$ for $fin M^*$, $xin M$, $alphain R$.
Now suppose that $R$ is commutative with a nontrivial involution and $M$ is a left $R$-module. We have two ways of taking the dual of $M$: one is by considering the natural right action of $R$ on $M$, which would give $falpha=alpha f$; the other is by considering the right action coming from twisting the left action via the involution, which would give $falpha=alpha^* f$. You can see that they are different (e.g. if $R=mathbb{C}$ and $alpha=i$ then the second action has a negative sign with respect to the first). So we have to be careful in this case, because we could be giving the same name to two different objects.
Edited to address further questions
1) We have $(Moplus N)^*=M^*oplus N^*$: by definition, if $fin(Moplus N)^*$ then $f:Moplus Nrightarrow R$. Consider the projections $f_M:Mrightarrow R$, $f_N:Nrightarrow R$ such that $f(m+n)=f_M(m)+f_N(n)$ for $min M, nin N$, and the right $R$-module structures are compatible. This shows that $(Moplus N)^*subseteq M^*oplus N^*$. Conversely, if $f_Min M^*$ and $f_Nin N^*$ then $f:=f_M+f_Nin (Moplus N)^*$ and all operations go well, so the two modules are the same.
2) Let us see that $Rcong R^*$ as right modules: We suppose that $R$ has identity. If the action was not twisted, we could use the regular representation $phi:Rrightarrow$Hom$(R,R)$, $phi(r)=phi_r$ such that $phi_r(x):=xr$, but due to the twisting caused by the involution we are going to consider $phi_r(x):=r^*x$.
First we see that $phi$ is a homomorphism of right $R$-modules. If $r,sin R$ then $phi_{r+s}(x)=(r+s)^*x=(r^*+s^*)x=r^*x+s^*x=phi_r(x)+phi_s(x)$ for all $xin R$ implies that $phi(r+s)=phi(r)+phi(s)$, and $phi_{sr}(x)=(sr)^*(x)=(r^*s^*)x=r^*(s^*x)=r^*phi_s(x)=phi_s(x)cdot r$ implies that $phi(sr)=phi(s)cdot r$.
Now wee see that $phi$ is injective: since it is additive it is enough tho show that $phi(r)=0$ implies $r=0$. This is true because $phi_r(x)=0$ for all $x$ implies $r^*x=0$ for all $x$, in particular $r^*=r^*1=0$, hence $r=0$.
Finally, we show that $phi$ is surjective: Given $fin$Hom$(R,R)$ note that $f(x)=f(1x)=f(1)x$, so $f$ is determined by $f(1)$; hence $phi(f(1)^*)=f$ since $phi_{f(1)^*}(x)=(f(1)^*)^*x=f(1)x=f(x)$ for all $xin R$.
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
$endgroup$
When we work over a commutative ring $R$ and $M$ is a left $R$-module, the dual module $M^*$ is considered to be Hom$(M,R)$ with right $R$-module structure.
If we now take $R$ to be not necessarily commutative, then $M^*=$Hom$(M,R)$ has, a priori, only a left $R$-module structure, given by $(alpha f)x=alpha(fx)$ for $fin M^*$, $xin M$, $alphain R$. We'd want to put a right $R$-module structure on $M^*$, and we can do when $R$ has an involution $*$, for then we can "twist" the left action to a right one, with the cost of changing the scalar to its conjugate via the involution: we define $$fcdotalpha:=alpha^* f,$$
so that $(falpha)(x)=alpha^*(fx)$ for $fin M^*$, $xin M$, $alphain R$.
Now suppose that $R$ is commutative with a nontrivial involution and $M$ is a left $R$-module. We have two ways of taking the dual of $M$: one is by considering the natural right action of $R$ on $M$, which would give $falpha=alpha f$; the other is by considering the right action coming from twisting the left action via the involution, which would give $falpha=alpha^* f$. You can see that they are different (e.g. if $R=mathbb{C}$ and $alpha=i$ then the second action has a negative sign with respect to the first). So we have to be careful in this case, because we could be giving the same name to two different objects.
Edited to address further questions
1) We have $(Moplus N)^*=M^*oplus N^*$: by definition, if $fin(Moplus N)^*$ then $f:Moplus Nrightarrow R$. Consider the projections $f_M:Mrightarrow R$, $f_N:Nrightarrow R$ such that $f(m+n)=f_M(m)+f_N(n)$ for $min M, nin N$, and the right $R$-module structures are compatible. This shows that $(Moplus N)^*subseteq M^*oplus N^*$. Conversely, if $f_Min M^*$ and $f_Nin N^*$ then $f:=f_M+f_Nin (Moplus N)^*$ and all operations go well, so the two modules are the same.
2) Let us see that $Rcong R^*$ as right modules: We suppose that $R$ has identity. If the action was not twisted, we could use the regular representation $phi:Rrightarrow$Hom$(R,R)$, $phi(r)=phi_r$ such that $phi_r(x):=xr$, but due to the twisting caused by the involution we are going to consider $phi_r(x):=r^*x$.
First we see that $phi$ is a homomorphism of right $R$-modules. If $r,sin R$ then $phi_{r+s}(x)=(r+s)^*x=(r^*+s^*)x=r^*x+s^*x=phi_r(x)+phi_s(x)$ for all $xin R$ implies that $phi(r+s)=phi(r)+phi(s)$, and $phi_{sr}(x)=(sr)^*(x)=(r^*s^*)x=r^*(s^*x)=r^*phi_s(x)=phi_s(x)cdot r$ implies that $phi(sr)=phi(s)cdot r$.
Now wee see that $phi$ is injective: since it is additive it is enough tho show that $phi(r)=0$ implies $r=0$. This is true because $phi_r(x)=0$ for all $x$ implies $r^*x=0$ for all $x$, in particular $r^*=r^*1=0$, hence $r=0$.
Finally, we show that $phi$ is surjective: Given $fin$Hom$(R,R)$ note that $f(x)=f(1x)=f(1)x$, so $f$ is determined by $f(1)$; hence $phi(f(1)^*)=f$ since $phi_{f(1)^*}(x)=(f(1)^*)^*x=f(1)x=f(x)$ for all $xin R$.
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
edited Jan 23 at 10:46
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
answered Jan 23 at 8:56
Jose BroxJose Brox
3,15711128
3,15711128
$begingroup$
@ninjahatori Please, see the update in my answer.
$endgroup$
– Jose Brox
Jan 23 at 10:46
$begingroup$
@ninjahatori OK, let's see if I can help you!
$endgroup$
– Jose Brox
Jan 28 at 11:23
add a comment |
$begingroup$
@ninjahatori Please, see the update in my answer.
$endgroup$
– Jose Brox
Jan 23 at 10:46
$begingroup$
@ninjahatori OK, let's see if I can help you!
$endgroup$
– Jose Brox
Jan 28 at 11:23
$begingroup$
@ninjahatori Please, see the update in my answer.
$endgroup$
– Jose Brox
Jan 23 at 10:46
$begingroup$
@ninjahatori Please, see the update in my answer.
$endgroup$
– Jose Brox
Jan 23 at 10:46
$begingroup$
@ninjahatori OK, let's see if I can help you!
$endgroup$
– Jose Brox
Jan 28 at 11:23
$begingroup$
@ninjahatori OK, let's see if I can help you!
$endgroup$
– Jose Brox
Jan 28 at 11:23
add a comment |
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