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Is there a name for the function?
Clash Royale CLAN TAG#URR8PPP
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Let $G$ be a group. Suppose $psi: G rightarrow mathbbC^*$ is a group homomorphism. The function $phi: G rightarrow (mathbbC,+)$ has the property:
$$ phi(g_1 g_2) = phi(g_1) + psi (g_1) phi(g_2),$$
for $g_1, g_2$ in $G$.
Is there a name for a function with this property? Note that if $psi (g) = 1$ for all $g in G$, then $phi$ is a homomorphism.
abstract-algebra complex-analysis
asked Jul 19 at 2:29
future_algebraist
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up vote
2
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Let $G$ be a group. Suppose $psi: G rightarrow mathbbC^*$ is a group homomorphism. The function $phi: G rightarrow (mathbbC,+)$ has the property:
$$ phi(g_1 g_2) = phi(g_1) + psi (g_1) phi(g_2),$$
for $g_1, g_2$ in $G$.
Is there a name for a function with this property? Note that if $psi (g) = 1$ for all $g in G$, then $phi$ is a homomorphism.
abstract-algebra complex-analysis
asked Jul 19 at 2:29
future_algebraist
134
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $G$ be a group. Suppose $psi: G rightarrow mathbbC^*$ is a group homomorphism. The function $phi: G rightarrow (mathbbC,+)$ has the property:
$$ phi(g_1 g_2) = phi(g_1) + psi (g_1) phi(g_2),$$
for $g_1, g_2$ in $G$.
Is there a name for a function with this property? Note that if $psi (g) = 1$ for all $g in G$, then $phi$ is a homomorphism.
abstract-algebra complex-analysis
asked Jul 19 at 2:29
future_algebraist
134
Let $G$ be a group. Suppose $psi: G rightarrow mathbbC^*$ is a group homomorphism. The function $phi: G rightarrow (mathbbC,+)$ has the property:
$$ phi(g_1 g_2) = phi(g_1) + psi (g_1) phi(g_2),$$
for $g_1, g_2$ in $G$.
Is there a name for a function with this property? Note that if $psi (g) = 1$ for all $g in G$, then $phi$ is a homomorphism.
abstract-algebra complex-analysis
asked Jul 19 at 2:29
future_algebraist
134
asked Jul 19 at 2:29
future_algebraist
134
asked Jul 19 at 2:29
future_algebraist
134
asked Jul 19 at 2:29
future_algebraist
134
134
add a comment |Â
add a comment |Â
1 Answer
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Think of $psi : G to mathbbC^times$ as defining a representation of the group $G$ on the $1$-dimensional complex vector space $mathbbC$. Then you can call $phi$ a $1$-cocycle on $G$ with coefficients in $psi$, or equivalently, a crossed homomorphism $G to mathbbC$ (with respect to the representation $psi : G to mathbbC^times$ on $mathbbC$).
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Think of $psi : G to mathbbC^times$ as defining a representation of the group $G$ on the $1$-dimensional complex vector space $mathbbC$. Then you can call $phi$ a $1$-cocycle on $G$ with coefficients in $psi$, or equivalently, a crossed homomorphism $G to mathbbC$ (with respect to the representation $psi : G to mathbbC^times$ on $mathbbC$).
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
add a comment |Â
up vote
1
down vote
accepted
Think of $psi : G to mathbbC^times$ as defining a representation of the group $G$ on the $1$-dimensional complex vector space $mathbbC$. Then you can call $phi$ a $1$-cocycle on $G$ with coefficients in $psi$, or equivalently, a crossed homomorphism $G to mathbbC$ (with respect to the representation $psi : G to mathbbC^times$ on $mathbbC$).
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Think of $psi : G to mathbbC^times$ as defining a representation of the group $G$ on the $1$-dimensional complex vector space $mathbbC$. Then you can call $phi$ a $1$-cocycle on $G$ with coefficients in $psi$, or equivalently, a crossed homomorphism $G to mathbbC$ (with respect to the representation $psi : G to mathbbC^times$ on $mathbbC$).
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
Think of $psi : G to mathbbC^times$ as defining a representation of the group $G$ on the $1$-dimensional complex vector space $mathbbC$. Then you can call $phi$ a $1$-cocycle on $G$ with coefficients in $psi$, or equivalently, a crossed homomorphism $G to mathbbC$ (with respect to the representation $psi : G to mathbbC^times$ on $mathbbC$).
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
answered Jul 19 at 7:38
Branimir Ćaćić
9,69221846
9,69221846
add a comment |Â
add a comment |Â
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