TITLE: If $\alpha,\beta\in C$ are algebraic over $Q$, show that $\alpha+\beta$ is a root of a quadratic polynomial in $Q[x]$ (with conditions).
QUESTION [4 upvotes]: I need to prove (b) to (c):
I have shown that $[Q(\alpha,\beta):Q(\alpha+\beta)][Q(\alpha+\beta):Q]=2$
But i cant proceed from there any suggestions?
what if b=-1?
Edit:
any hints on how to proceed from c to a would also be appreciated

REPLY [3 votes]: By assumption we have $\alpha+\beta=a+b\beta+\beta=a+(b+1)\beta$.
Since $\beta$ is algebraic over $\mathbb{Q}$ with degree 2. It exists a polynomial
$f\in\mathbb{Q}[X]$ with $f(\beta)=0$ and $\deg(f)=2$.
From this you can construct a polynomial $g$ with $g(a+(b+1)\beta)=0$ with $\deg(g)=2$.


TITLE: Conditions on $\beta$ under which the trace pairing restricted to $\mathfrak{so}(V,\beta)$ is positive (negative) definite
QUESTION [1 upvotes]: Let $V$ be a finite dimensional vector space over $ \mathbb{R}$. Let
\begin{equation} \left\langle\:,\:\right\rangle:\mbox{End}(V)\otimes\mbox{End}(V)\rightarrow \mathbb{R}\end{equation}
denote the pairing given by $\left\langle A,B\right\rangle = \mbox{Tr}(AB)$. We know that this pairing is $\mbox{GL}(V)$-invariant, symmetric, non-degenerate, but not definite (i,e. neither positive definite nor  negative definite).
Supposed that $\beta:V\otimes V \rightarrow \mathbb{R}$ is a non-degenerate symmetric pairing. Recall $\mathfrak{so}(V,\beta)\subset \mathfrak{gl}(V)$ is defined by
$$\mathfrak{so}(V,\beta)=\left\{A\in\mathfrak{gl}(V)\:|\: \forall v,w\in V \: \beta(Av,w)+\beta(v,Aw)=0\right\}.$$
We know that $\mathfrak{so}(V,\beta)$ is a Lie subalgebra of $\mathfrak{gl}(V)$.
The question: Establish conditions on $\beta$ under which the restriction of the trace pairing (i,e. $\left\langle\:,\:\right\rangle$ restricted to $\mathfrak{so}(V,\beta)$) is positive (also conditions to be negative) definite.
My attempt: We know that given  $\beta:V\times V\rightarrow \mathbb{R}$ a non-degenerate symmetric form, for all $A\in \mbox{End}(V)$ there exist a unique $B\in \mbox{End}(V)$ which satisfies $\beta(Av,w)=\beta(v,Bw)$ for all $v,w\in V$.
The unique $B$ as above will be denoted by $A^{\dagger}$ and called the transpose of $A$ (with respect to $\beta$).
The assignment $A\rightarrow A^{\dagger}$ defines a linear map $(\cdot)^{\dagger}\mbox{End}(V)\rightarrow \mbox{End}(V)$ called transposition  (with respect to $\beta$).
Therefore, we can consider
$$\mathfrak{so}(V,\beta)=\left\{A\in\mathfrak{gl}(V)\:|\: A=-A^{\dagger} \: (\mbox{transpose with respect to }\beta)\right\}.$$
In this sense, we have

Restriction the trace  pairing is positive definite if $\mbox{Tr}(A^{\dagger}A)<$ for all $A\in \mathfrak{so}(V,\beta)$ with $A\neq 0$.
Restriction the trace  pairing is negative definite if $\mbox{Tr}(A^{\dagger}A)>$ for all $A\in \mathfrak{so}(V,\beta)$ with $A\neq 0$.

My question: Conditions 1  and 2 depend on $\beta$. Does the condition 1 and 2 answer the question? Can I characterize $A^{\dagger}$ in terms of $\beta$ of a better way? Are there more specific conditions on $\beta$ to answer the question?

REPLY [0 votes]: Yes to all three questions. By diagonalisation of forms, WLOG, $\beta$ is diagonal with $\pm 1$ on the main diagonal. Now let us just compute (and you can get an explicit formula for your transposition as well).
If $\beta$ is positive or negative definite, $so(V,\beta)$ is the set of skew-symmetric matrices. The trace form is negative definite as easily observed.
If $\dim V =2$ and $\beta$ is indefinite, $so(V,\beta)$ is one-dimensional. The trace form is positive definite as easily observed.
If $\dim V >2$ and $\beta$ is indefinite, you look at a subalgebra of $so(V,\beta)$, of matrices that are zero outside a diagonal $3\times 3$-minor, which you choose so that $\beta$ has both $+1$ and $-1$ in the corresponding positions. You can easily write two matrices $A$ and $B$ in this subalgebra such that $Tr(A^2)>0<Tr(B^2)$. Thus, the trace form  is indefinite.


TITLE: Hawking Temperature of the BTZ Black Hole
QUESTION [4 upvotes]: The metric of the BTZ Black Hole is given by
$$
ds^2 = - N^2 dt^2 + N^{-2} dr^2 +r^2(d\phi + N^\phi dt)^2
$$
with
$$
N^2 = -M+ \frac{r^2}{l^2} + \frac{J^2}{4 r^2}, \ \ \  \ \ \  N^\phi = -\frac{J}{2r}
$$
The $g_{rr}$ component of the metric is singular at points where $N^2=0$, yielding the horizons $r_\pm$
$$
r_\pm = \sqrt{ \frac{Ml^2}{2}\left( 1 \pm \sqrt{1-\left(\frac{J}{Ml}\right)^2} \right)}
$$
Then for these $r_\pm$, the metric component $g_{tt}$ does not vanish but becomes
$$
g_{tt} =\frac{J^2}{4r_\pm^2}
$$
Now the perscription I learned to find a Hawking Temperature at a horizon (e.g. Schwarzschild BH) is you expand the Wick rotated metric ($t\to i\tau$) around the solution where $g_{\tau\tau}$ vanishes, find the metric is actually flat at this point, and impose $\tau$-periodicity such that there is no conical singularity at the horizon. This period is then the inverse Hawking temperature $T_H^{-1}$ .
I don't see any singularities in the $g_{\tau\tau}$ component right now so no conical singularity will appear, and I don't know how to interpret this. Does this mean there is no restriction on $T_H$ and the $\tau$ periodicity is free? Or is my way to calculate it somehow not applicable to BTZ horizons?

REPLY [3 votes]: I think I found a coordinate transformation that shines more light on this. Instead of looking for a conical singularity in $g_{tt}$ alone I must look at conical singularities in any part of the metric (apart from $g_{rr}$) First of all the metric can be expressed in terms of $r_\pm$ and Wick rotated $t\to it_E$ such that
$$
ds_{E}^2 = \frac{ (r^2-r_+^2)(r^2+r_-^2)}{l^2r}dt_E^2 + \frac{l^2 r^2}{(r^2-r_+^2)(r^2+r_-^2)} dr^2 + r^2(d\phi + \frac{i r_+ r_-}{l r^2} dt_E)^2
$$
which under the coordinate transformation
$$
t'_E = r_+ t_E + r_-\phi, \ \ \ \phi' = r_+ \phi - r_-t_E, \ \ \ r'^2= \frac{r^2-r_+^2}{r_+^2+r_-^2}
$$
becomes
$$
ds_E^2 = \frac{r'^2}{l^2} dt_E'^2 +\frac{l^2}{1+r'^2} dr'^2 + (1+r'^2)d\phi'^2
$$
which for $r\to r_+$  ($r'\to 0$) becomes
$$
ds_E^2 = r'^2 dt_E'^2+dr'^2 +d\phi'^2
$$
which represents flat polar coordinates iff $t_E' \sim t_E' +2\pi$. Furthermore $\phi'$ is not periodic. So the periodicity of $t_E'$ is $\Delta t_E'=2\pi$, and of $\phi'$ is $\Delta \phi'=0$. Combining this yields
$$
2\pi = r_+ \Delta t_E + r_-\Delta \phi, \ \ \ \ 0 = r_+\Delta \phi - r_- \Delta t_E
$$
So the time periodicity becomes $\beta =\Delta t_E = \frac{2\pi l r_+}{r_+^2+r_-^2}$ whilst also setting a fixed periodicity for $\phi$. Obviously the Hawking temperature is then
$$
T_H = \frac{r_+^2+r_-^2}{2\pi l r_+}
$$