twisted Poisson structures, degenerate metrics and integrability properties of (2,0)-tensors

Question

Given a regular (constant rank) bi-vector $\Pi \in \Gamma(\bigwedge^2TM)$ on a smooth manifold $M$ the necessary and sufficient condition for the image of $\Pi^\sharp:T^*M\to TM$ to be an integrable distribution is that $\Pi$ is a twisted Poisson tensor with respect to a closed 3-form $\phi$, i.e. there exist a closed 3-form $\phi$ such that $$[\Pi,\Pi]=\Pi^\sharp(\phi)$$ (see e.g. arXiv:1104.0880).

What is the corresponding condition for a degenerate regular symmetric (2,0)-tensor (a degenerate metric) $g\in \Gamma(S^2TM)$ to produce an integrable distribution as the image of $g^\sharp:T^*M\to TM$?

And what about any regular (constant rank) tensor $T\in \Gamma(TM\otimes TM)$, with no special symmetry properties?

Answer 1

Perhaps you will find this a useful answer, at least for the symmetric case: Let $S^2_r(TM)\subset S^2(TM)$ denote the subbundle consisting of the cometrics of rank $r$. (This bundle is the disjoint union of $r{+}1$ smooth subbundles that are distinguished by the algebraic type of the symmetric bivector. However, this 'type division' won't play a role in this analysis.)

There is a natural first-order differential operator $$ \delta_r:C^\infty\bigl(S^2_r(TM)\bigr)\longrightarrow C^\infty\bigl(\Lambda^2(TM)\otimes \Lambda^r(TM)\otimes \Lambda^{r+1}(TM)\bigr) $$ with the property that $\delta_r(g)=0$ if and only if the $r$-rank subbundle $g^\sharp(T^\ast M)\subset TM$ is integrable.

This differential operator is defined as follows: Locally, $g$ can be written in the form $$ g = \sum_{i,j=1}^rg^{ij}\ X_iX_j\ , $$ where $X_1,\ldots, X_r$ are linearly independent vector fields, $g^{ij}=g^{ji}$ are functions, and $\Delta = \det (g^{ij})$ is nonvanishing. Now define (using the summation convention) $$ \delta_r(g) = \Delta\ g^{ik}g^{jl}\ \left(X_i{\wedge}X_j\ \otimes\ X_1{\wedge}X_2{\wedge}\cdots{\wedge}X_r \ \otimes\ [X_k,X_l]{\wedge}X_1{\wedge}X_2{\wedge}\cdots{\wedge}X_r\right). $$ It is easy to verify that this is independent of the choice of basis $X_i$, and so it is globally defined. It is also obvious that it vanishes if and only if the bundle $g^\sharp(T^\ast M)$ (which is spanned locally by the $X_i$) is integrable.

Note that $\delta_r$ vanishes identically unless $2 \leq r < n$, as would be expected. Also, $\delta_r(\lambda g) = \lambda^{r+2}\ \delta_r(g)$ for all nonvanishing functions $\lambda$.

If I had considered oriented cometrics of rank $r$ (i.e., I had, in addition, fixed an orientation of $g^\sharp(T^\ast M)$), then I could then have got rid of the $\Lambda^r(TM)$ factor in the definition of $\delta_r$. However, I didn't want to impose the orientability condition, and the $\Lambda^r(TM)$ factor fixes that problem.

The almost Poisson case: Although you already have a criterion in the almost Poisson case, note that there is a similar construction to test the integrability of the bundle $\Pi^\sharp(T^\ast M)$ in the case of an almost Poisson structure $\Pi$ of constant half-rank $r$: Let $\Lambda^2_r(TM)\subset \Lambda^2(TM)$ be the subbundle of bivectors of half-rank $r$, i.e., its sections are those $\Pi$ such that $\Pi^r$ is nonvanishing as a section of $\Lambda^{2r}(TM)$ while $\Pi^{r+1}\equiv0$.

There is a natural first-order differential operator $$ \partial_r:C^\infty\bigl(\Lambda^2_r(TM)\bigr)\longrightarrow C^\infty\bigl(\Lambda^2(TM)\otimes\Lambda^{2r+1}(TM)\bigr) $$ defined as follows: Locally, write $\Pi$ in the form $$ \Pi = \sum_{i,j=1}^{2r} {\tfrac12}\ a^{ij}\ X_i\wedge X_j\ , $$ where $X_1,\ldots,X_{2r}$ are linearly independent vector fields, $a^{ij}=-a^{ji}$ are functions, and $\mathrm{Pf}(a^{ij})$ is nonvanishing (since $\Pi^r = r!\ \mathrm{Pf}(a^{ij})\ X_1\wedge X_2\wedge\cdots\wedge X_{2r}\not=0$). Now define (again using the summation convention) $$ \partial_r(\Pi) = {\tfrac14}\ a^{ik}a^{jl}\ X_k{\wedge}X_l\otimes\ [X_i, X_j]{\wedge}\Pi^r\ . $$ It is easy to verify that this does not depend on the choice of basis $X_i$, and so it is well-defined.

It is evident that that $\partial_r(\Pi)$ vanishes if and only if the bundle $\Pi^\sharp(T^\ast M)$ (which is spanned locally by the $X_i$) is integrable. Also, $\partial_r(\lambda\ \Pi)= \lambda^{r+2}\partial_r(\Pi)$ for all nonvanishing functions $\lambda$.

This is a somewhat more explicit test that the one you proposed, I think.

The general case: It turns out that one does not need any assumption of symmetry or skew-symmetry to get the integrability tensor, just an assumption of constant rank. Let $\otimes^2_r(TM)\subset TM\otimes TM$ denote the subbundle consisting of the tensors of rank $r$. A section $\phi\in C^\infty\bigl(\otimes^2_r(TM)\bigr)$ can be written locally in the form $$ \phi = \sum_{i,j=1}^r f^{ij}\ X_i\otimes Y_j $$ where $X_1,\ldots,X_r$ are linearly independent vector fields, $Y_1,\ldots,Y_r$ are linearly independent vector fields, and $f=(f^{ij})$ is an invertible matrix, i.e., $\det(f)\not=0$. The vector fields $X_i$ are local sections of a globally defined bundle $\lambda_\phi\subset TM$ of rank $r$, and the vector fields $Y_i$ are local sections of a globally defined bundle $\rho_\phi\subset TM$ of rank $r$.

There is a natural first-order differential operator $$ D_r: C^\infty\bigl(\otimes^2_r(TM)\bigr)\longrightarrow C^\infty\bigl(\Lambda^2(TM)\otimes \Lambda^r(TM)\otimes \Lambda^{r+1}(TM)\bigr) $$ which, relative to the local expression given above, takes the form $$ D_r(\phi) = \det(f)\ f^{ik}f^{jl}\ \left(Y_k{\wedge}Y_l\ \otimes \ Y_1{\wedge}Y_2{\wedge}\cdots{\wedge}Y_r \ \otimes\ [X_i,X_j]{\wedge}X_1{\wedge}X_2{\wedge}\cdots{\wedge}X_r\right). $$ This operator has the property that $D_r(\phi)=0$ if and only if $\lambda_\phi$ is integrable.

Note that there is a well-defined involution $\iota$ on $C^\infty\bigl(\otimes^2_r(TM)\bigr)$ such that $$ \iota(\phi) = \sum_{i,j=1}^r f^{ij}\ Y_j\otimes X_i\ , $$ and the integrability of $\rho_\phi$ is equivalent to the equation $D_r\bigl(\iota(\phi)\bigr)=0$.

Answer 2

This is also only for the symmetric case.

Let $S(TM)=\bigoplus_k S^k(TM)$ and $\Gamma(S(TM))$ be the commutative graded algebra of symmetric purely contravariant tensor fields. This is the algebra of functions on $T^*M$ which are polynomial on each fiber, and it is invariant under the Poisson bracket on $T^*M$. This is the symmetric Schouten bracket. See the following paper for more information:

Michel Dubois-Violette, Peter W. Michor: A common generalization of the Frölicher-Nijenhuis bracket and the Schouten bracket for symmetric multi vector fields, Indagationes Math. N. S. 6 (1995), 51--66. (pdf)

It might be that one can just carry over the considerations from $\Gamma(\Lambda (TM))$ to $\Gamma(S(TM))$ using this bracket.

EDIT: The above did not work. So let me try again. I do not need symmetry now.

Suppose that $a\in\Gamma(TM\otimes TM)$ is of constant rank when viewed as $a:T^\star M\to TM$. Choose a Riemann metric $g$ on $M$ and consider $A=a\circ g:TM\to T^\star M\to TM$ which is again of constant rank. So $\ker(A)$ and $A(TM)$ are subbundles. Choose an isomorphism $B:TM\to TM$ such that $A.B$ equals the $g$-orthogonal projection $P$ onto $A(TM)$. $P$ can be canonically constructed from $g$ and $a$, up to some free choice of an isomorphism between $Im(A)^\bot$ and $\ker(A)$.
Now the Frölicher-Nijenhuis bracket of the vector valued 1-form $P$ is of the form $$ [P,P] = 2R + 2\bar R,\quad R(X,Y)= P[(Id-P)X,(Id-P)Y], \quad \bar R(X,Y)=(Id-P)[PX,PY] $$
$R$, the curvature, is exacly the obstruction against integrability of the kernel of $P$.
$\bar R$, the cocurvature, is exactly the obstruction against integrability of the image of $P$, which we are interested in.
So the condition is $\bar R=0$ which can be written as $$ 0=P^*[P,P]=[P,P].(P\times P):\Lambda^2TM \to TM. $$

Answer 3

i might have a partial answer, at least for the symmetric case. however my reasoning might not be completely stringent, so i still would like to read a better argument.

Let $X,Y\in \Gamma(Image(g^{\sharp}))$, then $X^i=g^{ik}\alpha_k$, $Y^i=g^{ik}\beta_k$ with $\alpha, \beta$ being any two 1-forms. Let us use Frobenius theorem to study the integrability of $Image(g^\sharp)\subset TM$ and compute: $$[X,Y]^j=X^k \frac{\partial Y^j}{\partial x^k}-Y^k \frac{\partial X^j}{\partial x^k} = $$ $$=g^{jl} g^{ks} (\alpha_s \frac{\partial \beta_l}{\partial x^k}-\beta_s \frac{\partial \alpha_l}{\partial x^k})+\alpha_p \beta_l (g^{pk}\frac{\partial g^{jl}}{\partial x^k}-g^{lk}\frac{\partial g^{jp}}{\partial x^k})=$$ $$=g^{jl} g^{ks} (\alpha_s \frac{\partial \beta_l}{\partial x^k}-\beta_s \frac{\partial \alpha_l}{\partial x^k})+\alpha_p \beta_l (\Gamma^{plj}-g^{jk}\frac{\partial g^{pl}}{\partial x^k})$$ Now this last expression is in the image of $g^\sharp$ if and only if the (completely contravariant) Christoffel symbol $\Gamma^{plj}$ factors as $\tilde{\Gamma}^{pl}_r g^{rj}$. Now, here I'm not quite certain, but doesn't this mean that the (contravariant) Levi-Civita connection of $g$ in $TM$ must reduce to a connection on the vector bundle $Image(g^\sharp)\subset TM$?