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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Irreducible function on Q[z]

Let f(z) \in \mathbb{Q}[z] be irreducible with degree n>1. If f has a root on the unit circle then n is even and

    \[z^n f\left( \frac{1}{z} \right) = f(z)\]

Solution

Let \alpha be a root of f with |\alpha|=1. Since f has real coefficients \bar{\alpha}= \frac{1}{\alpha} is also a root of f. The product z^n f\left( \frac{1}{z} \right) is a polynomial in \mathbb{Q}[z] of degree n ( its leading coefficient is f(0) ) with root \alpha. By the irreducibility of f we have

(1)   \begin{equation*} z^n f\left( \frac{1}{z} \right) =  c f(z) \end{equation*}

for some non zero rational number c. Setting z=1 we have that f(1)=c f(1). Since f(1) \neq 0  , by our hypotheses , c=1 hence z^n f\left( \frac{1}{z} \right) = f(z) . Setting z=-1 we get that f(-1)=(-1)^n f(-1) and because f(-1) \neq 0 we deduce that n is even.

Note: The above tells us that f(z) can be expressed in terms of z + \frac{1}{z}.

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Inequality with roots

Let a, b, c be positive real numbers. Prove that

    \[\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2\]

Solution

We apply the AM – GM inequality, thus:

    \begin{align*} \frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}} &=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}} \\ &\geq 6\sqrt[6]{\frac{a}{b}\frac{b}{4c}\frac{c}{27a}} \\ &=\frac{6}{\sqrt[6]{4\cdot 27}} \end{align*}

Hence it suffices to prove that 3>\sqrt[6]{4\cdot 27} which holds because it is equivalent to 3^6> 4\cdot 27.

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On permutation

Let x_1, x_2, \dots, x_k be vectors of m – dimensional Euclidean space such that x_1+x_2+\cdots+x_k=0. Prove that there exists a permutation \pi of the integers \{1, 2, \dots,k\} such that

    \[\left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|\leq\left ( \sum_{i=1}^{k} \left \| x_i \right \|^2 \right )^{1/2}\]

for each n=1,2, \dots, k.

Solution

We define \pi inductively. Set \pi(1)=1.Assume \pi is defined for i=1, 2, \dots, n and also

(1)   \begin{equation*} \left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 \end{equation*}

Note that (1) is true for n=1. We choose \pi(n+1) in a way that (1) is fulfilled for n+1 instead of n. Set y=\sum \limits_{i=1}^{n} x_{\pi(i)} and A=\{1,2, \dots, k\} \setminus \{\pi(i) : i=1,2, \dots, n\}. Assume that (y, x_r)>0 for all r \in A.  Then \left ( y , \sum \limits_{r \in A} x_r \right )>0 and in view of y + \sum \limits_{r \in A} x_r =0 ones gets -(y, y)>0 which is impossible. Hence , there is r\in A such that

(2)   \begin{equation*} (y, x_r)\leq 0\end{equation*}

Put \pi(n+1) = r . Then using (2) and (1) we have

    \begin{align*} \left \| \sum_{i=1}^{n+1} x_{\pi(i)} \right \|^2 &= \left \| y + x_r \right \|^2 \\ &=\left \| y \right \|^2 + 2\left ( y, x_r \right ) + \left \| x_r \right \|^2 \\ &\leq \left \| y \right \|^2 + \left \| x_r \right \|^2 \\ &\leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 + \left \| x_r \right \|^2\\ &= \sum_{i=1}^{n+1} \left \| x_{\pi(i)} \right \|^2 \end{align*}

which verifies (1) for n+1. Thus we define \pi for every n=1, 2, \dots, k. Finally from (1) we get

    \[\left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{k} \left \| x_i \right \|^2\]

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