Let us consider the domain $$ \Omega=\mathbb{C}\setminus (]-\infty, 0]\,\cup\,[1,+\infty[) $$ (the doubly cleft plane). On it, we have the functions, $z^{\alpha}(1-z)^{\beta}$ for $\alpha,\beta\in \mathbb{C}$. We want to find a basis of the $\mathbb{C}$ - linear span of these fonctions (which form a monoid) denoted here $\mathcal{A}$. $$ \mathcal{A}=span_\mathbb{C}\{z^{\alpha}(1-z)^{\beta}\}_{\alpha,\beta\in \mathbb{C}} $$

Beginning with $\alpha,\beta\in \mathbb{R}$ and using the four relations

  1. $\frac{z}{1-z}=\frac{1}{1-z}-1$ (first quadrant)
  2. $\frac{1-z}{z}=\frac{1}{z}-1$ (second quadrant)
  3. $1=(1-z)+z$ (third quadrant)
  4. $z=1-(1-z)$ (fourth quadrant)

one can prove that the set $\{z^{\alpha}(1-z)^{\beta}\}_{(\alpha,\beta)\in D}$ where $D$ is the domain $$ D=\{(\alpha,\beta)\in \mathbb{R}^2\,|\, 0\leq \alpha<1\}=[0,1[\times \mathbb{R} $$ is (linearly) generating $\mathcal{A}$. On the basis of examples, I smell that it is a basis of it but cannot prove that they are linearly independent.

Q1) Does anybody have a hint for proving that $\{z^{\alpha}(1-z)^{\beta}\}_{(\alpha,\beta)\in D}$ are $\mathbb{R}$-linearly independent ? $\mathbb{C}$-linearly independent ?

Q2) Stronger question : find a similar "fundamental domain" for $\alpha,\beta\in \mathbb{C}$.

  • $\begingroup$ generating what ? / a basis of what ? $\endgroup$
    – mercio
    Mar 15, 2016 at 22:45
  • $\begingroup$ @mercio The algebra of the title, I was not very explicit, sorry. I'll fix this. $\endgroup$ Mar 15, 2016 at 22:47

2 Answers 2


Since $\mathcal{A}$ lies in the ring of holomorphic functions on $\Omega$ (I will call this ring $C^\omega(\Omega)$), it is important to understand what does $z^\alpha$ means for $\alpha \in \mathbb{C}$. The meaning is clear when we set $z^\alpha=e^{ log(z)\alpha}$, and similarly $(1-z)^\beta = e^{log(1-z)\beta}$. Clearly, because of the choice of the region $\Omega$, the function $log$ used above can be well-defined. This choice of a $log$ function makes us wish to understand what happens when we make a change to this choice.

$\mathcal{A}\subset C^\omega(\Omega)$ is a $\mathbb{C}$-vector space, with a generating set given by $\{z^{\alpha}(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$. It is certainly clear that this generating set is not linearly independent. For a while, assume that it is possible to define the following $\mathbb{C}$-linear functions $T_{x_i}:\mathcal{A}\rightarrow \mathcal{A},\ (i=0,1)$ such that the following happens $$ T_{x_0}:z^\alpha(1-z)^\beta \mapsto z^\alpha(1-z)^\beta e^{(2\pi i)\alpha} $$ $$ T_{x_1}:z^\alpha(1-z)^\beta \mapsto z^\alpha(1-z)^\beta e^{(2\pi i)\beta} $$

Now, we should be worried if this linear map is well-defined. For instance, if an element is written in two different ways as a $\mathbb{C}$-linear combination of elements $\{z^{\alpha}(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$, would these operators defined on this generating set yield consistent output? I will postpone this concern to when I define the maps $T_{x_0}, T_{x_1}$ more geometrically after justifying their importance.

Let us first see the benefit that these operators will have. First of all, each element of our generating set $\{z^{\alpha}(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$ is an eigenvector of this linear map. Now note the following lemma.

Lemma: For a $\mathbb{C}$-vector space $V$ and a map $T:V\rightarrow V$, a finite collection of eigenvectors corresponding to distinct eigenvalues are linearly independent over $\mathbb{C}$.

Proof can be found here

Now suppose we have a set $F \subset \mathbb{C}^2$ which contains a linear dependency relation of the following type ($c_{\alpha,\beta} \in \mathbb{C}$, $G \subset F$ is a non-empty finite set)

$$ \sum_{(\alpha,\beta) \in G} c_{\alpha,\beta}\ z^\alpha (1-z)^\beta =0 \ \ \ \ \ ...(1) $$

Without loss of generality, we assume that the choice of $G$ is such that $G$ contains the minimum possible number of terms. This leads us to believe that all the terms of the type $z^\alpha (1-z)^\beta$ in the above sum must reside in the same eigenspace of $T_{x_i},\ (i=0,1)$. Indeed, because if this were not the case, there would be elements of at least two different eigenspaces in the above sum and it should be possible to segregate the terms to get eigenvectors of different eigenspaces. Each of those individual sums will then have to be zero by the previous lemma causing a contradiction of the minimality of $G$.

Since the eigenvalues are the same, we assume that it is $e^{(2\pi i)a}$ for $T_{x_0}$ and $e^{(2\pi i)b}$ for $T_{x_1}$ for some $a,b \in \mathbb{C}$. This tells us that in the above sum, each $z^\alpha (1-z)^\beta$ is actually of the form $z^{a+m} (1-z)^{b+n}$ for some $m,n \in \mathbb{Z}$. We promptly divide the sum by $z^a (1-z)^b$ to modify our relation (1) and get the following $$ \sum_{(m+a,n+b) \in G} c_{(m+a),(n+b)}\ z^m (1-z)^n = 0 $$

More importantly, $G$ is just a translation of some set $G' \in \mathbb{Z}^2$ via $(a,b) \in \mathbb{C}^2$. So with the above discussion, we arrive at the following lemma.

Lemma: For $F \subset \mathbb{C}^2$, $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ can be linearly dependent over $\mathbb{C}$ (i.e. there is a finite subset of $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ that is linearly dependent over $\mathbb{C}$) if and only if there exists a finite set $G' \subset \mathbb{Z}^2$ such that $\{ z^m (1-z)^n \}_{(m,n) \in G'}$ are linearly dependent over $\mathbb{C}$ and $G'$ can be translated to become a subset of $F$ (i.e. there exists $(a,b)\in \mathbb{C}^2$ such that $(m+a,n+b) \in F,\ \forall (m,n) \in G'$).

Proof: If $G'$ can be translated to fit inside $F$ then clearly we are done (multiplying our linear dependence relation inside $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in G' }$ by a suitable $z^a(1-z)^b$ gives us a dependence relation in $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ ). Moreover, by the previous discussion, all minimal linear dependencies in the set $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ can only be of the above type.

We conclude our discussion with the following Proposition, which will give us a way to creating subsets $D \subset \mathbb{C}^2$ such that $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$ becomes a basis of $\mathcal{A}$.

Proposition: Suppose $A\subset \mathbb{Z}^2$ is such that $\{z^m (1-z)^n\}_{(m,n)\in A}$ is a basis of the vector space $\mathbb{C}\{z^m(1-z)^n\}_{(m,n)\in \mathbb{Z}^2} \subset C^\omega(\Omega)$, then $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$ is a basis of the vector space $\mathbb{C}\{z^\alpha(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$ where $D\subset \mathbb{C}^2$ is given by $$ D = \{ (m+a,n+b)|\ (m,n) \in A,\ a,b \in \mathbb{C},\ 0\leq \Re(a) < 1,\ 0\leq \Re(b) < 1 \} $$

Proof: Let us denote the eigenspace of $T_{x_i}$ corresponding to the eigenvalue $e^{(2\pi i )\alpha}$ by $V^i_{\alpha}$ ($i=0,1$). Let $W_{\alpha,\beta} = V_\alpha^0 \cap V_\beta^1 $. Clearly, $W_{0,0}\cap \mathcal{A} = \mathbb{C}\{z^m(1-z)^n\}_{(m,n)\in \mathbb{Z}^2}$. All elements of $W_{0,0}\cap\mathcal{A}$ can therefore be generated by $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$. Similarly, for $a,b \in \mathbb{C},\ 0\leq \Re(a) < 1,\ 0\leq \Re(b) < 1$ all elements of $W_{a,b}\cap \mathcal{A}$ can be generated by $\{z^{(m+a)} (1-z)^{(n+b)}\}_{(m,n)\in A}$. This way, we see that $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$ is a generating set of $\mathcal{A}$.

The fact that it is a basis can be established by the previous lemma. If D had a linearly dependence inside it, then we should be able to find a minimal linearly dependent set within some $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D} \cap W_{a,b}$ for some $a,b \in \mathbb{C}$. But, this set cannot contain any linear dependence by the choice of A.

One such $D$ can be constructed using $A=\{(-m,0),m\in \mathbb{N}\cup\{0\}\}\cup (\{0\}\times \mathbb{Z})$.

So what are these operators $T_{x_0}, T_{x_1}$?

In what follows, I will establish how the mysterious operators $T_{x_i}:\mathcal{A}\rightarrow \mathcal{A}$ emerge. By the virtue of the construction, it will become clear that the operators are well-defined linear functions on $\mathcal{A}$

We define $B=\mathbb{C}\backslash \{0,1\}$. Thus, we have a inclusion $\Omega \hookrightarrow B$. Let $p: E\rightarrow B$ be the universal covering space of $B$ such that $e \in E$ is mapped to $p(e)=b \in \Omega \subset B$. Since $\Omega$ is simply connected, the inclusion $\Omega \hookrightarrow B$ can be lifted to $E$ such that $\Omega \hookrightarrow E$ maps $b$ to $e$.

$$\require{AMScd} \begin{CD} \Omega @>\text{inclusion}>> E\\ @| @VVV \\ \Omega @>\text{inclusion}>> B \end{CD}$$

It is well known that $E$ is also a Complex $1$-manifold (a Riemann surface) like $\Omega$ and $B$ and it is possible to define holomorphic functions of $E$. Moreover, we have a natural action of the fundamental group $\pi_1(B,b)$ on $E$ as holomorphic deck transformations ( each $g \in \pi_1(B,b)$induces a biholomorphic homeomorphism $t_g:E\rightarrow E$ such that $p \circ t_g = p$ ). Hence we have the following diagram

$$\require{AMScd} \begin{CD} \Omega @>\text{inclusion}>> E @>t_g>> E\\ @| @VpVV @VpVV \\ \Omega @>\text{inclusion}>> B @= B\end{CD} $$

It is also well-known that the $\pi_1(B,b)$ is the free group generated by two generators. We choose two generators $x_0, x_1 \in \pi_1(B,b)$ that correspond to the path homotopy class of paths with endpoints at $b\in B$ and going around $0, 1$ in clockwise and anti-clockwise fashion respectively.

Corresponding to the diagram above, we have the opposite diagram induced on the ring of holomorphic functions on each of the Riemann surfaces

$$\require{AMScd} \begin{CD} C^\omega(\Omega) @<\text{Res}^{E}_{\Omega}<< C^\omega(E) @<T_g<< C^\omega(E) \\ @| @Ap_*AA @Ap_*AA \\ C^\omega(\Omega) @<\text{Res}^{B}_{\Omega}<< C^\omega(B) @= C^\omega(B)\end{CD} $$.

It is important to note that since $\Omega$ is embedded as open subset of $E$, the restriction map $\text{Res}^E_\Omega$ is injective. Morevoer, $\mathcal{A} \subset \text{Res}^E_\Omega(C^\omega(E))$. We can embed $\mathcal{A}$ into $C^\omega(E)$ by the following way. The following inclusion when right-composed to the restriction will give an identity map on $\mathcal{A}$

$$ \mathcal{A} \hookrightarrow C^\omega(E) $$ $$ z^\alpha (1-z)^\beta \mapsto e^{\alpha \mathbf{log_0}} e^{\beta \mathbf{log_1}} $$

Here $\mathbf{log_0}:E\rightarrow \mathbb{C}$ and $\mathbf{log_1}:E\rightarrow \mathbb{C}$ are holomorphic functions such that $\text{Res}^{E}_{\Omega}(\mathbf{log_0})(z) = log(z)$ and $\text{Res}^{E}_{\Omega}(\mathbf{log_1})(z) = log(1-z)$, for $z\in \mathbb{C}$ and the $log$ function is what was selected originally to define $\mathcal{A}$. Of course, we need to show that $\mathbf{log_0}, \mathbf{log_1}$ must exist. This detail can be given on request.

With the setup above, $T_{x_0}$ and $T_{x_1}$ have a clear meaning as linear functions on $C^\omega(\Omega)$ when we put $g=x_0,x_1$. For the purpose of the above discussion, we will have to restrict these functions to $\mathcal{A}$ where the inclusion $\mathcal{A} \hookrightarrow C^\omega(E)$ is as defined above. It can be easily checked that they give the output as desired. For instance, $T_{x_0}(e^{\alpha \mathbf{log_0}} e^{\beta \mathbf{log_1}} )(z) =e^{\alpha (\mathbf{log_0}(z)+2\pi i)} e^{\beta \mathbf{log_1}(z)}$, for $z \in E$ and the output is what is needed in the above discussion follows.

Hence with this, I finish my answer. Please let me know if there is a need for greater clarity anywhere in particular (especially in the places where I use the word "well-known").

  • $\begingroup$ I am now reading your paragraph So what are these operators $T_{x_0},T_{x_1}$ $\endgroup$ Aug 8, 2017 at 8:37
  • $\begingroup$ There may be some typographical errors in the answer. I am correcting them wherever I can spot. $\endgroup$ Aug 8, 2017 at 10:11
  • $\begingroup$ Thanks, I'm now testing the spaces and the arrows. It seems that one has to restrict $\Re(\beta)$ too. Am I right ? $\endgroup$ Aug 8, 2017 at 17:44
  • $\begingroup$ I am sorry, I do not understand. When you say "one has to restrict ℜ(β) too", which part are you referring to? $\endgroup$ Aug 9, 2017 at 9:21
  • $\begingroup$ I am so sorry about a confusion in the notation. The $B$ used before the "So what are these operators Tx0,Tx1"-paragraph has nothing to do with the $B$ used after it. I will fix this confusion in the notation. I will fix this. EDIT: I have fixed this by changing the first occurance of $B$ to $D$. $\endgroup$ Aug 9, 2017 at 9:47

During the week-end, I had the answer and proof to question 2 (which implies question 1). I post it considering that it can be useful to the community. If, however, somebody has references or sees a more elegant proof, I would appreciate and choose his answer.

In fact, one has the following

Proposition - The family \begin{equation} \{z^\alpha (1-z)^{\beta}\}_{(0\leq \Re(\beta)<1)\cup(0\leq \Re(\alpha)<1\,\wedge\, 0\leq \Re(\beta))} \end{equation} is a basis of the algebra \begin{equation} span_\mathbb{C} \{z^\alpha (1-z)^{\beta}\}_{\alpha,\beta\in \mathbb{C} } \end{equation}

First a function $g_{\alpha,\beta}=z^\alpha (1-z)^{\beta}$ ($\alpha,\beta\in \mathbb{C} $) being given, one writes $\alpha=k+\alpha_1$ with $k=\lfloor{\Re(\alpha)}\rfloor$. Then $0\leq \Re(\alpha_1)<1$ and \begin{eqnarray} z^\alpha (1-z)^{\beta}&=&z^{\alpha_1}z^{k}(1-z)^{\beta} =z^{\alpha_1}(1-(1-z))^{k}(1-z)^{\beta}\cr &=& \sum_{j=0}^k (-1)^j\binom{k}{j} z^{\alpha_1}(1-z)^{j+\beta} \end{eqnarray} this shows that \begin{equation} \{z^\alpha (1-z)^{\beta}\}_{0\leq\Re(\alpha)<1} \end{equation} is (linearly) generating. Remains to show that it is free. We need a lemma.

Lemma - Let $0<R<1$ and $\mathcal{R}_{0,R}$ be the ring of functions which are analytic at zero with convergence radius $R$ (i.e. $f\in \mathcal{R}_{0,R}$ iff $f(z)=\sum_{n\geq 0} a_n\,z^n$ for all $|z|< R$). Then, the set $\{z^{\alpha}\}_{0\leq\Re(\alpha)<1}$ is free over $\mathcal{R}_{0,R}$.

Note All functions are defined over $\Omega \cap B(0,R)$ where $B(0,R)$ is the open disk of radius $R$ around $0$ and $\Omega$ is the doubly cleft plane as in the question.

Proof - Let us consider linear relations of type \begin{equation} \sum_{i=1}^N z^{\alpha_i}S_i=0 \end{equation} on $\Omega \cap B(0,R)$ with $N\geq 1$, all $\alpha_i$ different and $S_i\not=0$. Then, either there is no one or there are and we will prove this second case to be impossible.

If there are relations of the preceding type, take one with $N$ minimal. We remark that $N\geq 2$ because there are no zero divisor in $\mathcal{H}(\Omega \cap B(0,R))$ because the domain is connected.

Using Euler's operator $\theta_0=z\frac{d}{dz}$, we get
\begin{eqnarray} 0&=&(\theta_0-\alpha_1Id)(\sum_{i=1}^N z^{\alpha_i}S_i)= \sum_{i=1}^N \Big((\alpha_i-\alpha_1)z^{\alpha_i}S_i+z^{\alpha_i}\theta_0(S_i)\Big)\cr &=&z^{\alpha_1}\theta_0(S_1)+\sum_{i=2}^N \Big((\alpha_i-\alpha_1)z^{\alpha_i}S_i+z^{\alpha_i}\theta_0(S_i)\Big) \end{eqnarray} and then \begin{eqnarray} 0&=&S_1(\theta_0-\alpha_1Id)(\sum_{i=1}^N z^{\alpha_i}S_i)- \theta_0(S_1)(\sum_{i=1}^N z^{\alpha_i}S_i)\cr &=&\sum_{i=2}^N z^{\alpha_i}\Big((\alpha_i-\alpha_1)S_1S_i+S_1\theta_0(S_i)-\theta_0(S_1)S_i\Big) \end{eqnarray} $N$ being minimal, all the coefficients are zero i.e. \begin{equation} \theta_0(S_1)=\Big((\alpha_i-\alpha_1)+\frac{\theta_0(S_i)}{S_i}\Big)S_1 \qquad (1) \end{equation} for $2\leq i\leq N$. With the logarithmic derivative \begin{equation} \Lambda(f)=\frac{\theta_0(f)}{zf} \end{equation} (defined here on $\mathcal{H}(\Omega\cap B(0,R))^{\times}$, the non-zero elements) with the properties $$ \Lambda(f_1f_2)=\Lambda(f_1)+\Lambda(f_2)\mbox{ and }\Lambda(f)=\Lambda(g)\Longrightarrow f=\lambda g, \lambda\in \mathbb{C} ^* $$
Equation (1) is at once rephrased as \begin{equation} \Lambda(S_1)=\frac{\alpha_i-\alpha_1}{z}+\Lambda(S_i)=\Lambda(z^{\alpha_i-\alpha_1})+\Lambda(S_i)= \Lambda(z^{\alpha_i-\alpha_1}S_i) \end{equation} in particular it exists a non-zero constant $\lambda$ such that \begin{equation} S_1=\lambda z^{\alpha_2-\alpha_1}S_2 \end{equation} recall that the valuation $S=\sum_{n\geq 0}a_nz^n$ is $$ val(S)=\min\{n\in \mathbb{N}\sqcup\{-\infty\}\,|\, a_n\not=0\} $$ and, for all, $S\not=0$ it is an integer. Then one can write \begin{equation} T=\sum_{n\geq 0}a_nz^n=\eta z^{\alpha} \qquad (2) \end{equation} with $T=\frac{S_1}{S_2};\alpha=\alpha_2-\alpha_1$ if $val(S_2)\leq val(S_1)$ and $T=\frac{S_2}{S_1};\alpha=\alpha_1-\alpha_2$ in the other case. Applying the Euler operator to Eq. (2) gives \begin{equation} \theta_0(T)=\sum_{n\geq 1}na_nz^n=\alpha\eta z^{\alpha} \end{equation} and, the fact that $|\Re(\alpha)|<1$ implies that $T=a_0\not=0$. Hence $S_i=\lambda_iS_1\not=0$ and Eq. (1) can be rewritten as \begin{equation} \sum_{i=1}^N \lambda_i\,z^{\alpha_i}=0 \end{equation} which implies that all $\lambda_i=0$ because $\theta_0(z^{\alpha_i})=\alpha_i\,z^{\alpha_i}$. A contradiction. $\square$

End of the proof of Proposition - Let \begin{equation} \sum_{(\alpha,\beta)\in F}c(\alpha,\beta)\,z^\alpha (1-z)^{\beta}=0 \end{equation} with $F\subset_{finite}\{(\alpha,\beta)\in \mathbb{C} ^2\,|\,0\leq\Re(\alpha)<1\}$, then \begin{equation} \sum_{\alpha\in pr_1(F)}z^\alpha \Big(\sum_{\beta\in pr_2(F)} c(\alpha,\beta)\,(1-z)^{\beta}\Big)=0 \end{equation} and the lemma, for $R=1$, implies that for all $\alpha\in pr_1(F)$, the function $$ \Big(\sum_{\beta\in pr_2(F)} c(\alpha,\beta)\,(1-z)^{\beta}\Big) $$ is zero on $\mathcal{H}(\Omega\cap B(0,1))$. But, now, one can use the Euler operator $\theta_1$ around $1$, $\theta_1=(1-z)\frac{d}{dz}$ and eigenfunctions $$ \theta_1((1-z)^{\beta})=-\beta\, (1-z)^{\beta} $$ to see that all $c(\alpha,\beta)$ are zero. $\square$

  • $\begingroup$ In the proof of the the proposition on the top, you seem to be assuming that $k $ is a postive integer. This may not be the case. $\endgroup$ Aug 10, 2017 at 6:12
  • $\begingroup$ I'll have a look. It seems that one correct "fundamental domain" should be $$(0\leq \Re(\beta)<1)\cup(0\leq \Re(\alpha)<1\mbox{ and } 0\leq \Re(\beta))$$ $\endgroup$ Aug 10, 2017 at 7:24
  • $\begingroup$ That may not be correct. For instance, $$ z^1(1-z)^0 + z^0(1-z)^1 - z^0(1-z)^0 = (z) + (1-z) - 1 = 0$$ is a linear-dependence lying among terms that lie in your domain. In the past, during our personal interactions, this domain worked for the collection $\{ \frac{z^\alpha }{(1-z)^\beta} \}_{(\alpha,\beta) \in \mathbb{C}^2}$. However, in the consideration of this question, we're dealing with $\{ z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in \mathbb{C}^2}$. It's just a notational confusion. Try $$ ( 0 \leq \Re(\alpha) < 1) \cup (0 \leq \Re(\beta) < 1\text{ and } \Re(\alpha) \leq 0 ) $$ $\endgroup$ Aug 10, 2017 at 8:58
  • $\begingroup$ No. The same counterexample holds ($ z + (1-z) -1 = 0 $). The domain $$(0 \leq \Re(\alpha) < 1) \cup (0 \leq \Re(\beta)< 1 \text{ and } \Re(\alpha)≤0)$$ comes from the example in my answer ( where I mention that "one such $D$ can be constructed using $A = \{ (−m, 0)|\ m \in \mathbb{N} \cap \{0\}\} \cup (\{0\}×\mathbb{Z} )$" as an example to the proposition of the answer). $\endgroup$ Aug 10, 2017 at 10:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .