I love the coordinate free definition of a tensor density of weight w as the tensorial product of a tensor and a scalar density (See the wonderful Lee introduction to smooth manifolds for instance) But I dont know how to obtain the far less nice covariant derivative of a density as defined in the Lovelock book) I would like to know a book or papers that work with densities thorougly from a coordinate free approach including covariant derivatives Thank you very much
1 Answers
I don’t know many references here, but I’ll outline what I can.
The scalar density bundle of weight $s$ (or perhaps $-s$…) $\mathscr{D}^s(M)$ on a smooth manifold $M$ is a rank-$1$ vector bundle which is globally trivializable (no topological assumptions needed here… but of course there is no natural choice of trivialization, just as a $1$-dimensional vector space doesn’t have a ‘special‘ basis). This is because there is a way to define ‘positive’ and ‘negative’ $s$-densities, and so you can use the convexity of $(0,\infty)$ and a partition of unity argument to construct an everywhere positive scalar density. Lee provides the definition for the $s=1$ case in his book and in this answer for general $s$.
If we fix a nowhere vanishing section $\rho$, then every other section can be written as $f\rho$ for some function $f:M\to\Bbb{R}$. So, in order to specify a linear connection $\nabla^{\mathscr{D}^s(M)}$ on the density bundle, you just need to fix one such $\rho$ and specify what you would like $\nabla^{\mathscr{D}^s(M)}\rho$ to be (…and since $\nabla_X\rho$ must be a function multiplied by $\rho$, and since this function depends $C^{\infty}(M)$-linearly on $X$, this amounts to specifying a $1$-form on $M$), because then by the Leibniz rule for connections, you can calculate \begin{align} \nabla^{\mathscr{D}^s(M)}(f\rho)=df\otimes \rho+f\nabla^{\mathscr{D}^s(M)}\rho. \end{align} Once you have this connection, you can take its tensor product to get connections on the $(k,l)$ tensor density bundles of weight $s$, $T^{k}_l(TM)\otimes \mathscr{D}^s(M)$.
In the specific case of a semi-Riemannian manifold $(M,g)$, we obviously use the Levi-Civita connection $\nabla$ on $TM$, and extend it to the various tensor bundles $T^k_l(TM)$ (still denoted $\nabla$). For the density bundle note that we have the Riemannian $s$-density $\rho_{g,s}$ which is nowhere-vanishing, and we shall define the connection $\nabla^{\mathscr{D}^s(M),g}$, henceforth denoted simply $\nabla$, on the $s$-density bundle by requiring $\nabla(\rho_{g,s}):=0$. This is certainly desirable since $\rho_{g,s}$ is defined using the metric only, and we would like our connection to preserve the metric. Thus, we now have a connection on $T^k_l(TM)\otimes \mathscr{D}^s(M)$ which is metric-compatible (in both senses: $\nabla g=0$ and $\nabla(\rho_{g,s})=0$).
Principal Bundle Approach
Another viewpoint (perhaps the ‘better’ one?) is to look at principal bundles and their associated vector bundles: let $(P,\pi,M,G)$ be a principal bundle over $M$ with group $G$ acting on the right, equipped with a principal connection $H$ (a choice of complement for the vertical subbundle $VP$ of $TP$). Let $V$ be a vector space for which $G$ acts linearly on $V$ from the left, and let $E=P\times^GV=(P\times V)/G$ be the associated vector bundle. I’ll spout out a bunch of stuff (most of which can be found in Kobayashi-Nomizu Volume I, or spread out over Dieudonne, Vol III-IV)
- there is a bijection between the set of $E$-valued $k$-forms on $M$ and the set of horizontal, $G$-equivariant $V$-valued $k$-forms on $P$. In particular for $k=0$, there is a bijection between smooth sections of the vector bundle $E$ and smooth $G$-equivariant functions $P\to V$.
- given an $E$-valued $k$-form $\omega$ on $M$, let $\tilde{\omega}$ be the associated horizontal, equivariant $V$-valued form on $P$. Then, the exterior covariant differentials $d^{\nabla}\omega$ and $D^H\widetilde{\omega}$ are ‘related’ in the above sense (i.e a certain square diagram commutes)
- Chain rule. Let $(P,\pi,M,G,H)$ be as above and let $W,X$ be any two (finite-dimensional real) Banach spaces, and let $f:P\to W$ be a smooth function and $\phi:\widetilde{W}\to X$ be a smooth function defined on an open set $\widetilde{W}\subset W$ containing the image of $f$. Then, for each $p\in P$, $D^H(\phi\circ f)_p=D\phi_{f(p)}\circ D^Hf_p$, where $D\phi_{f(p)}$ denotes the usual Frechet derivative.
- Corollary: if $D^Hf=0$, then $D^H(\phi\circ f)=0$. In other words, if a function $f$ is ‘principal-covariantly constant’, then so is every function of $f$.
Let $M$ be a smooth $n$-dimensional manifold, and $F(M)$ the frame bundle (a principal $\text{GL}(n)$ bundle) equipped with some principal connection $H$ and let $\nabla$ be the associated linear connection on $TM$ (or equivalently, you can start with a linear connection on $TM$ and induce a principal connection on the frame bundle $F(M)$).
- by taking $V=T^k_l(\Bbb{R}^n)$ and the usual representation of $\text{GL}(n)$, the associated vector bundle is isomorphic to the $(k,l)$ tensor bundle $T^k_l(TM)$.
- By (1) from before, a $(k,l)$ tensor field, $Z$, on $M$ is equivalent to a smooth $G$-equivariant function $\zeta:F(M)\to T^k_l(\Bbb{R}^n)$, and $\nabla Z$ corresponds via (1) to $D^H\zeta$. In particular, a $(0,2)$ tensor field $g$ on $M$ is equivalent to an equivariant function $\gamma:F(M)\to M_{n\times n}(\Bbb{R})$, and $\nabla g$ corresponds to $D^H\gamma$, so $\nabla g=0$ if and only if $D^H\gamma=0$. In particular if $g$ is a semi-Riemannian metric on the base manifold $M$ and $\nabla$ is a metric-compatible linear connection on $TM$, then the induced principal connection satisfies $D^H\gamma=0$, and so by the corollary above, for any function $\phi:\text{GL}(n,\Bbb{R})\to X$ for any other real Banach space $X$, we have $D^H(\phi\circ \gamma)=0$ (note $g$ is non-degenerate so $\gamma$ takes values in the invertible matrices, so the composition makes sense).
- For the next example, let $V=\Bbb{R},s\in\Bbb{R}$ and consider the linear representation \begin{align} \text{GL}(n)\to\text{GL}(\Bbb{R})\cong\Bbb{R}\setminus\{0\},\qquad A\mapsto\frac{1}{|\det A|^s}\text{id}_{\Bbb{R}}\cong \frac{1}{|\det A|^s}.\tag{$*$} \end{align} The associated vector bundle in this case is the density bundle of weight $s$ (or perhaps it’s $-s$…), and is isomorphic to the one in the above link so I’ll continue to denote it as $\mathscr{D}^s(M)$. In particular, since this is an associated vector bundle to our original principal bundle $F(M)$, the principal connection $H$ induces a linear connection $\nabla^{H,\mathscr{D}^s(M)}$.
By taking tensor products of the connections (of the tensor bundle and the $s$-density bundle), you are now able to take covariant derivatives of tensor densities of arbitrary weight $s$.
Equivalence of both Approaches.
Let us now prove the equivalence of the two approaches. Let $(M,g)$ be a semi-Riemannian manifold. As mentioned above, the tensor field $g$ gives rise to an equivariant function $\gamma:F(M)\to GL(n,\Bbb{R})\subset M_{n\times n}(\Bbb{R})$ defined explicitly as follows: for each frame (ordered basis) $\mathbf{e}=(e_1,\dots, e_n)$ of the tangent space $T_{\pi(\mathbf{e})}M$, \begin{align} \gamma(\mathbf{e})&:=\bigg(g(e_i,e_j)\bigg). \end{align} (Note also that in this formalism, you can make sense of $\gamma_{ij}$ as a well-defined function on the principal frame bundle, even though $g_{ij}$ only makes sense once you choose a chart, or some local frame).
Then, the function $F(M)\to\Bbb{R}$ which ‘corresponds’ to the $s$-density $\rho_{g,s}$ is the function $f=|\det\gamma|^{s/2}$ (you can check this has the correct equivariance property). This is a smooth composition with $\gamma$, so by the corollary, we have $D^Hf=0$, which means the linear connection $\nabla^{H,\mathscr{D}^s(M)}$ induced by the principal connection annihilates $\rho_{g,s}$, just as we demanded initially.
A Final Computational Remark
If we denote the map in $(*)$ as $\beta$, then differentiating at the identity and using the known fact that the derivative of determinant at the identity is the trace functional, we see that $D\beta_I:\text{End}(\Bbb{R}^n)\to\text{End}(\Bbb{R})$ is given by \begin{align} D\beta_I(\xi)&=-s\frac{1}{|\det(I+0\cdot\xi)|^{s+1}}\cdot\text{tr}(\xi)\cdot\text{id}_{\Bbb{R}}=-s\cdot\text{tr}(\xi)\cdot\text{id}_{\Bbb{R}}, \end{align} i.e as a bilinear map it is $\beta’:\text{End}(\Bbb{R}^n)\times\Bbb{R}\to\Bbb{R}$, given by $\beta’(\xi,t)= -s\cdot\text{tr}(\xi)\cdot t$.
Next, we have the following theorem:
Let $(P,\pi,M,G)$ be a principal bundle with a principal connection with connection 1-form $\omega$, and let $\beta:G\to\text{GL}(V)$ be a linear representation of $G$ on a real finite-dimensional Banach space $V$, and let $\beta’:\mathfrak{g}\times V\to V$ be the bilinear map as induced above. Then, for every $V$-valued $k$-form $\Phi$ on $P$ which is horizontal and $G$-equivariant, we can compute the exterior covariant differential by the formula \begin{align} D\Phi&=d\Phi+\omega\wedge_{\beta’}\Phi, \end{align} where $\wedge_{\beta’}$ is wedge product relative to the bilinear pairing $\beta’$.
If you now apply this formula to our specific case above, you see that for every horizontal and $\text{GL}(n)$-equivariant $k$-form $\Phi$ on the frame bundle $F(M)$ (taking values in $\Bbb{R}$), we have that \begin{align} D\Phi&=d\Phi-s\cdot\text{tr}(\omega)\wedge \Phi, \end{align} where $\omega$ is the $\mathfrak{g}=\text{End}(\Bbb{R}^n)$-valued connection 1-form on $F(M)$, so taking its trace gives a usual 1-form on $F(M)$, and we can now take its wedge product with the usual $k$-form $\Phi$, i.e this is all basic stuff. In particular, if you specialize to $k=0$ (i.e $\Phi:F(M)\to\Bbb{R}$ is just an equivariant function) and to weight $s=1$, then (keep in mind now wedge product becomes basic scalar multiplication) this becomes \begin{align} D\Phi&=d\Phi-\text{tr}(\omega)\cdot\Phi. \end{align} The reason I bring this formula up is because if you look in other places e.g this PhySE answer, or Lovelock-Rund’s book (Chapter 4, eq. 1.23) you will find traced Christoffel symbols $\Gamma^i_{j\,i}$. So, the trace comes, geometrically/‘abstractly’, from differentiating the determinant which built-into the definition of the density bundle, and then simply working out how the single connection in the principal bundle propagates to the various associated vector bundles.
- 65,833
-
2Thank you very much for your answer. I have to work out all the information that you have provided but it seems very interesting to me. – Ismael cortazar Dec 20 '23 at 09:56