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edited May 15 2011 at 13:24 |
This is too elementary a question for MO. In the case of complex characters, the
kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$
On the other hand, if the character $\chi$ is afforded by a representation $\sigma$
(that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$,
the eigenvalues of $g \sigma$ are all $o(g)$-th roots of unity, where $o(g)$ is the order
of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way
we can have $\chi(g) = \chi(1)$ is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$
has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$
is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$
is clear.
Added later, in light of the discussion below: ``kernel" implicitly refers to the kernel of a group homomorphism here (actually, several group homomorphisms, as we will see). The (complex)
character $\chi$ may be afforded by several different homomomorphisms homomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C})).$ Hence all such representations
have the same kernel in the group-theoretic sense. Furthermore, the argument above shows that the kernel can be seen directly from the character values. Hence the kernel of a character is really the kernel of an equivalence class of representations. One of the (many) advantages of working with (complex) characters is that all normal subgroups of a finite group $G$ can be determined by inspection of the character table of $G$.
It would also be possible to speak of the kernel of an algebra homomorphim, but that would be the set of all elements of the group algebra $\mathbb{C}G$ sent to the zero matrix by a chosen representation $\sigma$, and we would need to consider the algebra homomorphism as mapping into the full matrix ring $M_{n}(\mathbb{C})$. However, this kernel is not so easy to read fom from the character values on group elements.
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edited May 15 2011 at 11:08 |
This is too elementary a question for MO. In the case of complex characters, the
kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$
On the other hand, if the character $\chi$ is afforded by a representation $\sigma$
(that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$,
the eigenvalues of $g \sigma$ are all $o(g)$-th roots of unity, where $o(g)$ is the order
of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way
we can have $\chi(g) = \chi(1)$ is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$
has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$
is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$
is clear.
Added later, in light of the discussion below: ``kernel" implicitly refers to the kernel of a group homomorphism here (actually, several group homomorphisms, as we will see). The (complex)
character $\chi$ may be afforded by several different homomomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C})).$ Hence all such representations
have the same kernel in the group-theoretic sense. Furthermore, the argument above shows that the kernel can be seen directly from the character values. Hence the kernel of a character is really the kernel of an equivalence class of representations. One of the (many) advantages of working with (complex) characters is that all normal subgroups of a finite group $G$ can be determined by inspection of the character table of $G$.
It would also be possible to speak of the kernel of a ring an algebra homomorphim, but that would be the set of all elements of the group algebra $\mathbb{C}G$ sent to the zero matrix by a chosen representation $\sigma$, and we would need to consider the ring algebra homomorphism as mapping into the full matrix ring $M_{n}(\mathbb{C})$. However, this kernel is not so easy to read fom the character values on group elements.
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edited May 15 2011 at 10:47 |
This is too elementary a question for MO. In the case of complex characters, the
kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$
On the other hand, if the character $\chi$ is afforded by a representation $\sigma$
(that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$,
the eigenvalues of $g \sigma$ are all $o(g)$-th roots of unity, where $o(g)$ is the order
of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way
we can have $\chi(g) = \chi(1)$ is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$
has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$
is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$
is clear.
Added later, in light of the discussion below: ``kernel" implicitly refers to the kernel of a group homomorphism here (actually, several group homomorphisms, as we will see). The (complex)
character $\chi$ may be afforded by several different homomomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C}).$ GL}(n,\mathbb{C})).$ Hence all such representations
have the same kernel in the group-theoretic sense. Furthermore, the argument above shows that the kernel can be seen directly from the character values. Hence the kernel of a character is really the kernel of an equivalence class of representations. One of the (many) advantages of working with (complex) characters is that all normal subgroups of a finite group $G$ can be determined by inspection of the character table of $G$.
It would also be possible to speak of the kernel of a ring homomorphim, but that would be the set of all elements of the group algebra $\mathbb{C}G$ sent to the zero matrix by a chosen representation
$\sigma$, and we would need to consider the ring homomorphism as mapping into the full matrix ring $M_{n}(\mathbb{C})$. However, this kernel is not so easy to read fom the character values on group elements.
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4
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edited May 15 2011 at 10:41 |
This is too elementary a question for MO. In the case of complex characters, the
kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$
On the other hand, if the character $\chi$ is afforded by a representation $\sigma$
(that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$,
the eigenvalues of $g \sigma$ are all $o(g)$-th roots of unity, where $o(g)$ is the order
of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way
we can have $\chi(g) = \chi(1)$ is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$
has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$
is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$
is clear.
Added later, in light of the discussion below: ``kernel" implicitly refers to the kernel of a \emph{group} group homomorphism here (actually, several group homomorphisms, as we will see). The (complex)
character $\chi$ may be afforded by several different homomomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C}).$ Hence all such representations
have the same kernel in the group-theoretic sense. Furthermore, the argument above shows that the kernel can be seen directly from the character values. Hence the kernel of a character is really the kernel of an equivalence class of representations. One of the (many) advantages of working with (complex) characters is that all normal subgroups of a finite group $G$ can be determined by inspection of the character table of $G$.
It would also be possible to speak of the kernel of a ring homomorphim, but that would be the set of all elements of the group algebra $\mathbb{C}G$ sent to the zero matrix by a chosen representation
$\sigma$. \sigma$, and we would need to consider the ring homomorphism as mapping into the full matrix ring $M_{n}(\mathbb{C})$. However, this kernel is not so easy to read fom the character values on group elements, and in any case, no group element is sent to the zero matrix.
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edited May 15 2011 at 10:33 |
This is too elementary a question for MO. In the case of complex characters, the
kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$
On the other hand, if the character $\chi$ is afforded by a representation $\sigma$
(that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$,
the eigenvalues of $g \sigma$ are all $o(g)$-th roots of unity, where $o(g)$ is the order
of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way
we can have $\chi(g) = \chi(1)$ is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$
has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$
is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$
is clear.
Added later, in light of the discussion below: ``kernel" implicitly refers to the kernel of a \emph{group} homomorphism here (actually, several group homomorphisms, as we will see). The (complex)
character $\chi$ may be afforded by several different homomomorphisms $\sigma: G \to {\rm GL}(n,\mathbb{C})$, where $n = \chi(1)$. But all such representations are equivalent( that is. detemined up to conjugation by a matrix in ${\rm GL}(n,\mathbb{C}).$ Hence all such representations
have the same kernel. Furthermore, the argument above shows that the kernel can be seen directly
from the character values. Hence the kernel of a character is really the kernel of an equivalence
class of representations.
It would also be possible to speak of the kernel of a ring homomorphim, but that would be the set of all elements of the group algebra $\mathbb{C}G$ sent to the zero matrix by a chosen representation
$\sigma$. However, this is not so easy to read fom the character values on group elements, and in any case, no group element is sent to the zero matrix.
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edited May 14 2011 at 22:12 |
This is too elementary a question for MO. In the case of complex characters, the
kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$
On the other hand, if the character $\chi$ is afforded by a representation $\sigma$
(that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$,
the eigenvalues of $g \sigma$ are all $o(g)$-the o(g)$-th roots of unity, where $o(g)$ is the order
of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way
we can have $\chi(g) = \chi(1)$ if is if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$
has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$
is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$
is clear.
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answered May 14 2011 at 20:25 |
This is too elementary a question for MO. In the case of complex characters, the
kernel of a character $\chi$ of a finite group $G$ is $\{g \in G: \chi(g) = \chi(1) \}.$
On the other hand, if the character $\chi$ is afforded by a representation $\sigma$
(that is, $\chi(g) = {\rm trace}(g\sigma)$ for all $g \in G$), then for each $g \in G$,
the eigenvalues of $g \sigma$ are all $o(g)$-the roots of unity, where $o(g)$ is the order
of $g$. Hence the triangle inequality yields $|\chi(g)| \leq \chi(1)$, and the only way
we can have $\chi(g) = \chi(1)$ if if all eigenvalues of $g\sigma$ are equal to 1. Since $g\sigma$
has finite order, in that case, $g \sigma$ must be the identity matrix. Thus ${\rm ker}\chi$
is precisely equal to ${\rm ker} \sigma$, as the inclusion ${\rm ker}\sigma \leq {\rm ker}\chi$
is clear.
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