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− | == Part a) == | + | ==Task== |
| + | We endow the Cartesian product \(G=SO\left(3\right)\times\mathbb{R}^{3}\) with the map \(\circ:G\times G\rightarrow G\) defined by \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\). |
| + | <ol style="list-style-type:lower-latin"> |
| + | <li>Check that the map \(\circ\) turns \(G\) into a group. Find explicit formulas for the neutral element and inverse.</li> |
| + | <li>Verify that the formula \(\left(\rho\left(A,y\right)f\right)\left(x\right):=f\left(A^{-1}\left(x-y\right)\right)\) defines a representation of \(G\) on the space of real valued functions \(f\) on \(\mathbb{R}^{3}\)</li> |
| + | </ol> |
| + | ==Solution== |
| + | ===Subtask a)=== |
| + | Check that \((A,x) \circ (B,y) \in G\): |
| + | \(SO(3)\) is a group under matrix multiplication, so \(AB \in SO(3)\). \(Ay, x \in \mathbb{R}^3\), and since \(\mathbb{R}^3\) is a vector space, \(Ay + x \in \mathbb{R}^3\). So, the statement is proven. |
| | | |
− | Prove that, if \(f\) is an elliptic function, then \(f\) is constant if and only if \(f\) has no poles.
| + | We check the associativity of the group multiplication: |
| + | $$ |
| + | \begin{align} |
| + | \left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)&=\left(AB,Ay+x\right)\circ\left(C,z\right)\\ |
| + | &=\left(\left(AB\right)C,ABz+Ay+x\right)\\ |
| + | &=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)\\ |
| + | &=\left(A,x\right)\circ\left(BC,Bz+y\right)\\ |
| + | &=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)\\ |
| + | \end{align} |
| + | $$ |
| + | The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\) because |
| + | $$(A,x) \circ (\mathbb{I},0) = (A\mathbb{I},A0 + x) = (A,x) = (\mathbb{I}A, \mathbb{I}x + 0) = (\mathbb{I},0) \circ (A,x)$$ and |
| + | $$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$ |
| | | |
− | === Solution part a) === | + | ===Subtask b)=== |
− | | + | We check that it is indeed a homomorphism: |
− | '''Definition''' (elliptic function): A function \(f\) is called elliptic if it has the following two properties:
| + | $$ |
− | | + | \begin{align} |
− | (a) \(f\) is doubly periodic.
| + | \left(\rho [ \left(A,y\right)\circ\left(B,z\right) ] f\right)\left(x\right) &= \left(\rho [ AB,Az+y ] f\right)\left(x\right) \\ |
− | | + | &= f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right) \\ |
− | (b) \(f\) is meromorphic (its only singularities in the finite plane are poles).
| + | &= f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right) \\ |
− | | + | &= f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right) \\ |
− | (Taken from Apostol, 1.4)
| + | &= \left(\rho\left(A,y\right)\left(f\left(B^{-1}\left(\cdot-z\right)\right)\right)\right)\left(x\right) \\ |
− | | + | &= \left(\rho\left(A,y\right)\rho\left(B,z\right)f\right)\left(x\right) \\ |
− | '''Liouville's theorem''': If \(f\) is holomorph on \( \mathbb{C} \) and bounded, then \(f\) is constant.
| + | |
− | | + | |
− | | + | |
− | '''To prove''': Let \(f\) be an elliptic function: \(f\) is constant \(\Leftrightarrow\) \(f\) has no poles.
| + | |
− | | + | |
− | \( "\Rightarrow" \) Let \(f\) be constant on \( \mathbb{C} \) \( \rightarrow f\) has no poles.
| + | |
− | | + | |
− | \( "\Leftarrow" \) Let \(P\) be the fundamental region to \(f\), where \(f\) is elliptic and has no poles. | + | |
− | | + | |
− | \( f(\mathbb{C}) = f(P) \) is compact and thus bounded. Because \(f\) is both holomorph and bounded on \(\mathbb{C}\), Liouville's theorem tells us that \(f\) has to be a constant function. \(\square\) | + | |
− | | + | |
− | | + | |
− | ==Part b)== | + | |
− | | + | |
− | \( \vdash: \) Let \( Q = \left[ 0, 1\right] \times \left[ 0, 1 \right] \subset \mathbb{C} \) be the unit square, and let \( f \) be a holomorphic function on a neighborhood of \( Q \). Suppose further that \( f( z + i) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, 1 \right] \) and \( f( z + 1) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, i \right] \). Then \(f\) is constant. | + | |
− | | + | |
− | | + | |
− | ===Proof 1===
| + | |
− | | + | |
− | We write \( z = x + iy \) and \( f(x + iy) = u(x,y) + i v(x,y) \) where \( u, v \) are real valued functions. By assumption we get:
| + | |
− | | + | |
− | $$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$
| + | |
− | | + | |
− | $$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$
| + | |
− | | + | |
− | Since \( f \) is a holomorphic function the integral \( \int_{\partial Q } f(z) \, dz = \int_{\partial Q } u + iv \, dz = 0 \). We divide \( \partial Q = Q_1 \cup Q_2 \cup Q_3 \cup Q_4 \) where \( Q_{1, ..., 4} \) are the edges of the square. We imagine \( Q_1 \) to be the interval \( [0, 1] \), and proceed to numerate the edges in positive orientation.
| + | |
− | | + | |
− | Then:
| + | |
− | | + | |
− | $$ \int_{\partial Q } u + iv \, dz = \int_{\partial Q } u(z) \, dz + i \int_{\partial Q } v(z) \, dz = \int_{\partial Q } u(z) \, dz + i \left( \int_{Q_1 } + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) v(x,y) \, dz $$
| + | |
− | | + | |
− | But since the function values of \( v \) on \( Q_1 \) are the same as on \( Q_3 \) and are the same on \( Q_2 \) as on \( Q_4 \) we get
| + | |
− | | + | |
− | $$ \int_{\partial Q } v(z) \, dz = 0 $$
| + | |
− | | + | |
− | because of the converse orientation of the \( Q_1 \) and \( Q_3 \) part, respectively the \( Q_2 \) and \( Q_4 \) part.
| + | |
− | | + | |
− | Thus we have \( \int_{\partial Q } u(z) \, dz = 0 \). We write, since \( \forall y \in \left[ 0, 1 \right] : u(1, y) = u(0, y) + \alpha (y) \) and \( \forall x \in \left[ 0, 1 \right] : u(x, 1) = u(x, 0) + \beta (x) \) where \( \alpha, \beta \in \mathbb{R}_{\geq 0} \):
| + | |
− | | + | |
− | $$ \begin{align}
| + | |
− | \int_{\partial Q } u(z) \, dz &= \int_{\partial Q } u(x, y) \, d(x + iy) \\
| + | |
− | &= \int_{\partial Q } u(x, y) \, dx + i \int_{\partial Q } u(x, y) \, dy \\
| + | |
− | &= \left( \int_{Q_1 } + \int_{Q_3} \right) u(x, y) \, dx + i \left(\int_{Q_2 } + \int_{Q_4 } \right) u(x, y) \, dy \\ | + | |
− | &= \int_{[0, 1]} u(x,0) \, dt - \int_{[0, 1]} u(x, 1) \, dt + i \left( \int_{[0, 1]} u(1, t) \, dt - \int_{[0, 1]} u(0,t) \, dt \right) \\
| + | |
− | &= \int_{[0, 1]} \left( - \alpha (t) \right) \, dt + i \left( \int_{[0, 1]} \beta(t) \, dt \right) = 0
| + | |
| \end{align}$$ | | \end{align}$$ |
| + | (Here the dot indicates where the argument of the function goes) |
| | | |
− | but since \( \alpha, \beta \) are non-negative it follows that \( \alpha \equiv 0 \) and \( \beta \equiv 0 \) because otherwise the integrals wouldn't disappear. We now can doubly-periodically and holomorphically continue \( f \) on \( \mathbb{C} \) to an elliptic function with the fundamental pair of periods \( \left( 1, i \right) \).
| + | We furthermore show trivially that \(\rho(e) = e\\ |
− | | + | \rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\) |
− | We can do this since the continuation is obviously continuous, and we have by the identity principle that \( f(z) \equiv f(z + i ) \) on a neighborhood of each \( z \in [0, 1] \) and \( f(z) \equiv f(z + 1) \) for each \( z \in [0, i] \) (''note'': the identity principle holds since every point \( z \in \partial Q \) is a limit point of the coincidence set of \( f ( z + i ) = f (z) \), respectively \( f( z + 1) = f(z) \)).
| + | |
− | | + | |
− | We then see easily that the limit \( f'(z) = \lim_{\Delta z \rightarrow 0} \frac{f(z + \Delta z ) - f(z)}{\Delta z} = \lim_{\Delta z \rightarrow 0} \frac{f(z + i + \Delta z ) - f(z + i)}{\Delta z} = f'(z + i) \) for \( z \in [0,1] \) and of course also \( f(z) = f(z + 1) \) for \( z \in [0, i] \) exists. Thus the continuation has an existing derivative of first order at all points in \( \mathbb{C} \) which is, by its periodicity on the boundary, also continuous, ergo the continuation of \( f \) is an entire function. But since \( f \) is entire and elliptic, it follows from part a) that \( f \equiv const \). \( \square \)
| + | |
− | | + | |
− | ===Proof 2===
| + | |
− | | + | |
− | ''You have to be careful here! You can't use Liouville's theorem without proof. If you want to state this proof in the exam you have to learn a proof for Liouville. For all I've seen this is a rather long and complicated proof. If you know a short and easy one, I guess we would all be glad if you stated it here. A.''
| + | |
− | | + | |
− | We write \( z = x + iy \) and \( f(x + iy) = u(x,y) + i v(x,y) \) where \( u, v \) are real valued functions. By assumption we get:
| + | |
− | | + | |
− | $$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$
| + | |
− | | + | |
− | $$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$
| + | |
− | | + | |
− | Since \( f \) is holomorphic, \( v \) is a harmonic function. We can harmonically and doubly-periodically continue \( v \) to \( v : \mathbb{C} \rightarrow \mathbb{R} \). Furthermore \( v(Q) = v(\mathbb{C}) \) and since \( Q \) is compact we have \( \forall z \in \mathbb{C} \left| v (z) \right| \leq M \) for some \( M \in \mathbb{R} \). We now apply Liouville's theorem for harmonic functions and get \( v \equiv const \).
| + | |
− | | + | |
− | From the Cauchy-Riemann equations it follows immediately that also \( u \) has to be constant. \( \square \)
| + | |
− | | + | |
− | '''Liouville's theorem for harmonic functions:''' Let \( \lambda : \mathbb{C} \rightarrow \mathbb{R} \) be a harmonic function and \( \forall z \in \mathbb{C} \left| \lambda (z) \right| \leq M \) for some \( M \in \mathbb{R} \). Then \( \lambda \) is constant.
| + | |
− | | + | |
− | ===Proof 3===
| + | |
− | | + | |
− | Let \( \partial Q\) be the path along the boundary of Q, from cauchy follows \( \int_{\partial Q } f(z) \, dz = 0 \)
| + | |
− | | + | |
− | We then split the Integral into Q1, Q2, Q3, Q4 as above, and parameterize them, it follows:
| + | |
− | | + | |
− | $$ \int_0^1 f(t) - f(i+t) dt + i \int_0^1 f(1+it) - f(it) dt = 0$$
| + | |
− | | + | |
− | from the exercise we know that both Integrants are real -> both Integrals are real -> both Integrals must be zero, because the real and imaginary part of the equation must be individually zero:
| + | |
− | | + | |
− | $$ \int_0^1 f(i+t) - f(t) dt = 0 , \int_0^1 f(1+it) - f(it) dt = 0$$
| + | |
− | | + | |
− | Where I multiplied the first equation by -1. Both Integrats are non-negative, it follows that:
| + | |
| | | |
− | $$f(i+t) = f(t) \ and \ f(1+it) = f(it) \ for \ all \ t \in \left[ 0, 1 \right] $$
| + | Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\) |
| | | |
− | It follows that f is double periodic and can be analyticly continued on \(\mathbb{C}\), from (a) follows that f is constant
| + | \(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\) |
− | <p style="text-align:right;">\(\square\)</p>
| + | |
We endow the Cartesian product \(G=SO\left(3\right)\times\mathbb{R}^{3}\) with the map \(\circ:G\times G\rightarrow G\) defined by \(\left(A,x\right)\circ\left(B,y\right):=\left(AB,Ay+x\right)\).
Check that \((A,x) \circ (B,y) \in G\):
\(SO(3)\) is a group under matrix multiplication, so \(AB \in SO(3)\). \(Ay, x \in \mathbb{R}^3\), and since \(\mathbb{R}^3\) is a vector space, \(Ay + x \in \mathbb{R}^3\). So, the statement is proven.
We check the associativity of the group multiplication:
$$
\begin{align}
\left(\left(A,x\right)\circ\left(B,y\right)\right)\circ\left(C,z\right)&=\left(AB,Ay+x\right)\circ\left(C,z\right)\\
&=\left(\left(AB\right)C,ABz+Ay+x\right)\\
&=\left(A\left(BC\right),A\left(Bz+y\right)+x\right)\\
&=\left(A,x\right)\circ\left(BC,Bz+y\right)\\
&=\left(A,x\right)\circ\left(\left(B,y\right)\circ\left(C,z\right)\right)\\
\end{align}
$$
The identity element is \(\left(\mathbb{I},0\right)\) and the inverse element is \(\left(A^{-1},-A^{-1}x\right)\) because
$$(A,x) \circ (\mathbb{I},0) = (A\mathbb{I},A0 + x) = (A,x) = (\mathbb{I}A, \mathbb{I}x + 0) = (\mathbb{I},0) \circ (A,x)$$ and
$$(A,x) \circ (A^{-1},-A^{-1}x) = (AA^{-1},-AA^{-1}x + x) = (\mathbb{I},0) =(A^{-1}A,A^{-1}x-A^{-1}x)=(A^{-1},-A^{-1}x) \circ (A,x)$$
We check that it is indeed a homomorphism:
$$
\begin{align}
\left(\rho [ \left(A,y\right)\circ\left(B,z\right) ] f\right)\left(x\right) &= \left(\rho [ AB,Az+y ] f\right)\left(x\right) \\
&= f\left(\left(AB\right)^{-1}\left(x-Az-y\right)\right) \\
&= f\left(B^{-1}A^{-1}\left(x-Az-y\right)\right) \\
&= f\left(B^{-1}\left(A^{-1}\left(x-y\right)-z\right)\right) \\
&= \left(\rho\left(A,y\right)\left(f\left(B^{-1}\left(\cdot-z\right)\right)\right)\right)\left(x\right) \\
&= \left(\rho\left(A,y\right)\rho\left(B,z\right)f\right)\left(x\right) \\
\end{align}$$
(Here the dot indicates where the argument of the function goes)
We furthermore show trivially that \(\rho(e) = e\\
\rho(e)(f)(x) = \rho((\mathbb{I},0))(f)(x) = f(\mathbb{I}(x-0)) = f(x)\)
Using the homomorphic property of \(\rho\) and the fact that \(G\) is a group: \(e = \rho(e) = \rho(gg^{-1}) = \rho(g) \rho(g^{-1}) \overset{!}{=} \rho(g) \rho(g)^{-1} \Rightarrow \rho(g) \in GL(V)\)
\(\Rightarrow \rho\) is a representation of \(G\) on \(V\) the space of all real vaued functions on \(\mathbb{R}^3\)