Group homomorphism of uncountable kernel

Let f:\mathbb{C}^* \rightarrow \mathbb{R}^* be a group homomorphism. We proved in this question that its kernel is infinite. We are proving now that it is also uncountable.

Solution

Before we proceed with the proof we are stating that not all homomorphisms are of the form z\mapsto |z|^\alpha. We can find non trivial homomorphisms. But all continuous are of the above form.

Let \mu be the group of roots of unity. Both groups \mathbb{R}^* / \mu(\mathbb{R}) and \mathbb{C}^* / \mu(\mathbb{C}) are uniquely divisible and thus are vector spaces over \mathbb{Q}.

Since the positive reals are closed under multiplication, it’s easy to see that

    \[\mathbb{R}^* \cong \mu(\mathbb{R}) \oplus \mathbb{R}^* / \mu(\mathbb{R})\]

Using the axiom of choice  ,  we   construct a group homomorphism

    \[\mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C}) \cong \mathbb{Q}^{\mathfrak{c}} \cong \mathbb{R}^* / \mu(\mathbb{R}) \to \mathbb{R}^*\]

Hence the kernel is uncountable.

Note: \mu(\mathbb{C}) is  a direct summand of \mathbb{C}^*. That is because

    \[\operatorname{Ext}(\mathbb{C}^*/\mu(\mathbb{C}), \mu(\mathbb{C})) \cong \operatorname{Ext}(\mathbb{Q}^{\mathfrak{c}}, \mathbb{Q}/\mathbb{Z}) \cong 0\]

( since \mathbb{Q} / \mathbb{Z} is injective )  and so the exact sequence

    \[0 \to \mu(\mathbb{C}) \to \mathbb{C}^* \to \mathbb{C}^* / \mu(\mathbb{C}) \to 0\]

splits.

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Group homomorphism with an infinite kernel

Let f:\mathbb{C}^* \rightarrow \mathbb{R}^* be a homomorphism. Prove that kernel of f is infinite.

Solution

If \mathcal{G} < \mathbb{C}^\ast is a finite subgroup then all z \in \mathcal{G} must have norm 1, i.e. z \in \mathbb{S}^1. ( Otherwise otherwise z, z^2, z^3, \dotsc is an infinite sequence of distinct elements in \mathcal{G}. )

Suppose , that \mathcal{G} is finite and is the kernel of f:\mathbb{C}^\ast \rightarrow \mathbb{R}^\ast. Then let z \in \mathcal{G} and

    \[f(2z) = f(2)f(z) = 2\cdot 0 = 0\]

So,  2z \in \ker f = G. But this is a contradiction since |2z| = 2 \neq 1.

Hence,  no finite subgroup \mathal{G} of \mathbb{C}^\ast can be a kernel.

Note: The homomorphisms that can easily be described are for the form f(z)=|z|^\alpha.

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Zero determinant of a matrix

Suppose that for the complex square matrices A, B it holds

(1)   \begin{equation*} AB - BA =A \end{equation*}

Prove that \det A =0.

Solution

If \det A\ne0, then A is invertible. So we get

    \[I=A^{-1}A=A^{-1}(AB-BA)=B-A^{-1}BA\]

But this equality is impossible: taking trace on both sides, we get

n=\text{Tr}(I)=\text{Tr}(B-A^{-1}BA)=\text{Tr}(B)-\text{Tr}(A^{-1}BA)=\text{Tr}(B)-\text{Tr}(B)=0

We get a contradiction, which shows that A is singular, i.e. \det A=0.

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On a power of matrix

Let n be a natural number such that n \geq 2. Evaluate the power

    \[\mathcal{P} = \begin{pmatrix} 1 &1 \\ 1&0 \end{pmatrix}^n\]

Solution

This is a very standard exercise in diagonalisation of matrices and there would be no reason to post it here , if it did not include the Fibonacci result. We are proving that

    \[\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}\]

where F_n denotes the n – th Fibonacci number. The proof now follows with an induction on n.

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Multiple integral on a zero measured set

Let A \subseteq \mathbb{R}^n be a Jordan measurable set of zero measure. Prove that \bigintsss \limits_{A} 1 \, {\rm d} \bar{x}= 0.

Solution

Since \mu(A)=0 there exists a sequence U_n of closed rectangles of \mathbb{R}^n such that A \subseteq \bigcup \limits_{n} U_n and \forall \epsilon>0 it is \sum \limits_{n} \mathcal{V} \left( U_n \right)< \epsilon where \mathcal{V}(U_n) is the volume of the rectangle U_n. Then foreach \epsilon>0

    \begin{align*} 0 &\leq \int \limits_{A} 1 \, {\rm d} \bar{x} \\ &\leq \int \limits_{\bigcup \limits_{n} U_n} 1 \, {\rm d} \bar{x} \\ &\leq \sum_{n} \int \limits_{U_n} 1\, {\rm d}\bar{x} \\ &= \sum_{n} \mathcal{V} \left ( U_n \right )\\ &< \epsilon \end{align*}

 

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