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Metric Tensor of Hyperboloid Model for Hyperbolic Space with Curvature $K$

1

$begingroup$

Let $mathbbH^d_K$ denote the set of points of the hyperboloid model, which models the hyperbolic space with curvature $K$ (where $K$ is negative). So

$$
mathbbH^d_K=_*=sqrtK
$$

where $|p|_*=sqrtlangle p,prangle_*$ and $langle p,prangle_*=-p_0p_0+p_1p_1+...+p_dp_d.$

Now, how do you derive the metric tensor $g$ for any curvature $K$?

---

What I know so far:

I know that when $K=-1$ we have $g=diag(-1,1,1,...,1)inmathbbR^(d+1)times (d+1)$. And we have that
$$
langle p,prangle_*=p^Tgp.
$$
But what if $Kneq -1$? How does $g_K$ then look like?

hyperbolic-geometry

share|cite|improve this question

edited Apr 2 at 18:46

ndrizza

asked Apr 2 at 17:52

ndrizzandrizza

510312

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  By $-K^1/2$ do you mean $(-K)^1/2$? Otherwise, with $K < 0$, things don't seem to make sense. . . Cheers!
  $endgroup$
  – Robert Lewis
  Apr 2 at 18:10
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes! I've updated it. Thank you!
  $endgroup$
  – ndrizza
  Apr 2 at 18:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you replace $p_alpha$ by $q_alpha=fracp_alpha(-K)^1/2$, then the equation for $H_K$ reduces to the equation for $H_1$. Thus, if you know the metric tensor for the latter, you can recover the metric tensor for the former.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 18:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GiuseppeNegro This means that if $g_K$ denotes the metric tensor for curvature $K$, we have that $g_K=fracpsqrtK$?
  $endgroup$
  – ndrizza
  Apr 2 at 18:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Terrible choice of notation; if $p_alpha$ denotes points then you should use $g_alphabeta$ or $h_alphabeta$ to denote metric tensor. Anyway, I think there should be a $|K|$, not a $sqrtK$, and I don't know if it is in the numerator or in the denominator.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 18:42
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

Let $mathbbH^d_K$ denote the set of points of the hyperboloid model, which models the hyperbolic space with curvature $K$ (where $K$ is negative). So

$$
mathbbH^d_K=_*=sqrtK
$$

where $|p|_*=sqrtlangle p,prangle_*$ and $langle p,prangle_*=-p_0p_0+p_1p_1+...+p_dp_d.$

Now, how do you derive the metric tensor $g$ for any curvature $K$?

---

What I know so far:

I know that when $K=-1$ we have $g=diag(-1,1,1,...,1)inmathbbR^(d+1)times (d+1)$. And we have that
$$
langle p,prangle_*=p^Tgp.
$$
But what if $Kneq -1$? How does $g_K$ then look like?

hyperbolic-geometry

share|cite|improve this question

edited Apr 2 at 18:46

ndrizza

asked Apr 2 at 17:52

ndrizzandrizza

510312

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  By $-K^1/2$ do you mean $(-K)^1/2$? Otherwise, with $K < 0$, things don't seem to make sense. . . Cheers!
  $endgroup$
  – Robert Lewis
  Apr 2 at 18:10
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes! I've updated it. Thank you!
  $endgroup$
  – ndrizza
  Apr 2 at 18:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you replace $p_alpha$ by $q_alpha=fracp_alpha(-K)^1/2$, then the equation for $H_K$ reduces to the equation for $H_1$. Thus, if you know the metric tensor for the latter, you can recover the metric tensor for the former.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 18:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GiuseppeNegro This means that if $g_K$ denotes the metric tensor for curvature $K$, we have that $g_K=fracpsqrtK$?
  $endgroup$
  – ndrizza
  Apr 2 at 18:41
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Terrible choice of notation; if $p_alpha$ denotes points then you should use $g_alphabeta$ or $h_alphabeta$ to denote metric tensor. Anyway, I think there should be a $|K|$, not a $sqrtK$, and I don't know if it is in the numerator or in the denominator.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 18:42
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Let $mathbbH^d_K$ denote the set of points of the hyperboloid model, which models the hyperbolic space with curvature $K$ (where $K$ is negative). So

$$
mathbbH^d_K=_*=sqrtK
$$

where $|p|_*=sqrtlangle p,prangle_*$ and $langle p,prangle_*=-p_0p_0+p_1p_1+...+p_dp_d.$

Now, how do you derive the metric tensor $g$ for any curvature $K$?

---

What I know so far:

I know that when $K=-1$ we have $g=diag(-1,1,1,...,1)inmathbbR^(d+1)times (d+1)$. And we have that
$$
langle p,prangle_*=p^Tgp.
$$
But what if $Kneq -1$? How does $g_K$ then look like?

hyperbolic-geometry

share|cite|improve this question

edited Apr 2 at 18:46

ndrizza

asked Apr 2 at 17:52

ndrizzandrizza

510312

$endgroup$

share|cite|improve this question

edited Apr 2 at 18:46

ndrizza

asked Apr 2 at 17:52

ndrizzandrizza

510312

share|cite|improve this question

edited Apr 2 at 18:46

ndrizza

asked Apr 2 at 17:52

ndrizzandrizza

510312

asked Apr 2 at 17:52

ndrizzandrizza

510312

asked Apr 2 at 17:52

ndrizzandrizza

510312