1
$\begingroup$

Both in Barton Zwiebach's A First Course In String Theory and R. Blumenhagen's Basic Concepts of String Theory, when the Nambu-Goto action $$S_{NG}=\int d^2\sigma \sqrt{-\det(\gamma_{ab})}$$ is presented, the argument to the fact that $\det(\gamma_{ab}) =\dot{X^2} X^{\prime 2} - (\dot{X} \cdot X^\prime)^2< 0$ involves requiring that a vector $v$ in the embedded worldsheet (embedded submanifold) written in the form $v = \dot{X} + \lambda X^\prime, \lambda \in \mathbb{R}$ must be time-like or space-like as $\lambda$ varies. Basically, making $v^2(\lambda) = f(\lambda)$ we get that in order $f(\lambda)$ assume both positive and negative vakues, the equation $f(\lambda)$ must have two real difference solutions, wich implies the discriminant $\Delta >0$ which gives the desired result.

My question: the above argument for Strings is totally doable, but unpractical for higher dimensions. Given a Lorentzian manifold $(M, g)$ and an embedded (by a smoth map $\phi: \Sigma \subset M \longrightarrow M$) submanifold with induced metric $(\Sigma, \gamma := \phi^{*}g)$, how can I show, in a different manner that I mentioned above, if $\gamma$ is Lorentzian or not?

$\endgroup$
2
  • $\begingroup$ Just to clarify, are you asking how - if presented with a metric manifold $(\mathcal M,g)$ and a generic submanifold $\Sigma \subset \mathcal M$ - one might determine the signature of the induced metric $\gamma$? $\endgroup$
    – J. Murray
    Sep 22, 2021 at 0:44
  • $\begingroup$ @J. Murray: yes, that's what I want to know. $\endgroup$ Sep 22, 2021 at 0:46

1 Answer 1

0
$\begingroup$

Here's an idea that looks like it generalizes only because it's Minkowski space (the whole thing seems like it will fail if we start allowing multiple linearly independent time-like directions in the $(-,-,..,+,+,..,+)$ metric at each point). I will do the $(+,-,..,-)$ metric, it should adapt to the $(-,+,...,+)$ metric, and you can try to adapt it to more formal 'embedding language':

In Euclidean space with the Euclidean inner product, we can define the determinant to be an alternating multi-linear map which acts on the standard basis $\{\hat{\mathbf{e}}_1,...,\hat{\mathbf{e}}_n \}$ as $$\det(\hat{\mathbf{e}}_1,...,\hat{\mathbf{e}}_n) = ||\hat{\mathbf{e}}_1|| ... ||\hat{\mathbf{e}}_n|| \cdot 1 = 1$$ Thus for example any surface in 3D Euclidean space $\mathbf{r} = \mathbf{r}(u,v)$, with a parametrization $(u,v)$ such that $\frac{\partial \mathbf{r}}{\partial u}$ and $\frac{\partial \mathbf{r}}{\partial v}$ are linearly independent tangent vectors, is such that at any point on the surface the quantities $d_u \mathbf{r} = \frac{\partial \mathbf{r}}{\partial u} du$ and $d_v \mathbf{r} = \frac{\partial \mathbf{r}}{\partial v} dv$ are just two vectors (with the dimensions of $\mathbf{r}$) and so the area between them in Euclidean space is given by the absolute value of their wedge product in the tangent space at that point which can be written as $$|d_u \mathbf{r} \wedge d_v \mathbf{r}| = \sqrt{\det\left(\frac{\partial \mathbf{r}}{\partial u},\frac{\partial \mathbf{r}}{\partial v}\right)}\ du\ dv$$ which becomes the surface area element for $du$ and $dv$ infinitesimal.

In Minkowski space with the $(+,-,...,-)$ metric and the Minkowski inner product, we can define the determinant to be an alternating multi-linear map which acts on the standard basis $\{\hat{\mathbf{e}}_0,...,\hat{\mathbf{e}}_{D-1} \}$ as $$\det(\hat{\mathbf{e}}_0,...,\hat{\mathbf{e}}_{D-1}) = ||\hat{\mathbf{e}}_0|| ... ||\hat{\mathbf{e}}_{D-1}|| \cdot 1 = (-1)^{D-1}.$$

In the $(+,-,...,-)$ metric, the distance between two points in spacec-time is time-like or light-like if $$ds^2 = dx_{\mu} dx^{\mu} = dt^2 - d\mathbf{x}^2 \geq 0.$$ Therefore the velocity for the path of a time-like or light-like particle satisfies $$\left\lVert\frac{d x}{dt}\right\rVert^2 = 1 - (\mathbf{v})^2 \geq 0.$$ We now treat $x^{\mu}$ as a function of parameters $(\tau,\sigma,\rho,...) = (\sigma^{\alpha}) = (\sigma^0,\sigma^i)$ such that the tangent vectors $$\frac{\partial x^{\mu}}{\partial \sigma^{\alpha}}$$ are linearly independent, i.e. the $(\sigma^{\alpha})$ parametrization actually parametrizes a brane 'surface' in space-time.

If we choose our parametrization such that $\tau = t$ then for a time-like or light-like particle there must exist at a fixed $\tau = t$ the string, brane, ... generates a curve, surface, ... via $\mathbf{x} = \mathbf{x}(t,\sigma^i)$ so that $$\left\lVert\frac{\partial x}{\partial \sigma^i}\right\rVert^2 = 0 - \left(\frac{\partial \mathbf{x}}{\partial \sigma^i}\right)^2 < 0$$ i.e. we must have $p$ space-like vectors.

Since the time-like vector $d_t x = \frac{\partial x}{\partial t} dt$ and $p$ light-like vectors $d_i x = \frac{\partial x}{\partial \sigma^i} d \sigma^i$ are just linearly independent tangent vectors in Minkowski space-time, we can consider the 'area' they span (in a suitably oriented tangent space) as $$|d_t x \wedge d_1 x \wedge ... \wedge d_i x| = \sqrt{(-1)^p \det\left(\frac{\partial x}{\partial t},\frac{\partial x}{\partial \sigma^1},...,\frac{\partial x}{\partial \sigma^i}\right)}\ d\tau\ d \sigma^1 ... d \sigma^i $$ Since this is invariant under reparametrizations, it should hold for all other 'coordinate systems' $\sigma^{\alpha}$ so that $$|d_0 x \wedge d_1 x \wedge ... \wedge d_i x| = \sqrt{(-1)^p \det\left(\frac{\partial x}{\partial \sigma^0},\frac{\partial x}{\partial \sigma^1},...,\frac{\partial x}{\partial \sigma^i}\right)}\ d \sigma^0\ d \sigma^1 ... d \sigma^i $$ is the surface area of a $p$-brane generating a $(p+1)$-surface such that nowhere does it move faster than the speed of light.

That $(-1)^p \det\left(\frac{\partial x}{\partial \sigma^0},\frac{\partial x}{\partial \sigma^1},...,\frac{\partial x}{\partial \sigma^i}\right) \geq 0$ is the necessary and sufficient condition for there to exist one time-like and $p$ space-like tangent vectors to the $(p+1)$-surface generated by the $p$-brane in $(+,-,...,-)$ Minkowski space is seen as follows.

If there is only $1$ light-like vector and $p$ space-like vectors, then by the above definition of a determinant in Minkowski space we see the term under the square root can only be non-negative, and since we can get to this form by a reparametrization it holds for a general parametrization.

If that thing under the square root is non-negative, then we consider two cases, positive or zero. If it is non-zero then it is positive which means the $p+1$ tangent vectors from $D \geq p+1$ space that we plug into it can't all be space-like otherwise it would be negative because we'd have $(-1)^{2p+1}$ times a positive number, which is negative. Since there is at most one linearly independent time-like vector in Minkowski space at this point we must have one time-like and $p$ space-like vectors. If it is zero then because the $\frac{\partial x}{\partial \sigma^{\alpha}}$ are linearly independent, and because $\frac{\partial x}{\partial \sigma^{i}} < 0$, it means the only way this thing could be zero is if the norm of one of the vectors we plug into it is zero, but this is only possible for a time-like vector, and only linearly independent one exists in our space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.