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?

  • $\begingroup$ for completeness let me add that, in the case of twisted Poisson structures, if $i_{\mathcal{O}}:\mathcal{O}\to M$ is one of the induced leaves of the foliation, then $$i_{\mathcal{O}}^*(\phi)=d\Omega_{\mathcal{O}}$$ where $\Omega_{\mathcal{O}}$ is the 2-form inverse to the restriction of $\Pi$ to the foliation. In case $\phi=0$ we are in the Poisson case and $\Omega_{\mathcal{O}}$ is the symplectic form on the symplectic leaf $\mathcal{O}$. $\endgroup$ – issoroloap Dec 27 '12 at 1:25

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 $D_r\bigl(\iota(\phi)\bigr)=0$.

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  • $\begingroup$ Thank you also for this reply. This is an extremely direct test indeed. $\endgroup$ – issoroloap Dec 30 '12 at 14:52
  • $\begingroup$ great, thank you Robert! one question though: what about the linearity properties of you differential operator? I see that it is homogeneous of dregree r+2 with respect to multiplication of $\phi$ by a constant (even a function). what about linear combinations of tensors $a\phi+b\varphi$? $\endgroup$ – issoroloap Dec 30 '12 at 18:04
  • $\begingroup$ let me elaborate: one cool thing about the equation [\Pi,\Pi]=\Pi^\sharp(\phi) for the integrability of antisymmetric case is that one can put $\Pi=\Pi_0+\epsilon \Pi_1$ and get a compatibility condition between $\Pi_0$ and $\Pi_1$ in order for $\Pi$ to generate an integrable distribution (this is the idea of bihamiltonian systems). In the fully Poisson case it reduces to $[\Pi_0,\Pi_1]=0$ of course, but if we start with a Poisson tensor $\Pi_0$ but perturb it to anything integrable (but still antisymmetric) we get a nice and explicit system of equations involving $\Pi_0$, $\Pi_1, \phi$. $\endgroup$ – issoroloap Dec 30 '12 at 18:26
  • $\begingroup$ @issoroloap: $D_r$ does not have linearity properties in any usual sense because the fibers of the bundle $\otimes^2_r(TM)$ don't have a linear structure. I guess you could ask what you could say if you had a linear combination $\Phi=a\phi+b\varphi$ with the property that this was a section of $\otimes^2_r(TM)$ for all $a$ and $b$ (or an open subset of such coefficients). Offhand, I don't know of any examples of this that don't keep the two bundles $\lambda_\Phi$ and $\rho_\Phi$ fixed (in which case, the 'superposition formula' is kind of trivial), but I imagine that there could be some. $\endgroup$ – Robert Bryant Dec 30 '12 at 18:47
  • $\begingroup$ Yes i meant precisely that. I'll think about it a little more, but this is already more than satisfactory. Thanks again. $\endgroup$ – issoroloap Dec 30 '12 at 19:43

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 $$ 0=P^*[P,P]=[P,P].(P\times P):\Lambda^2TM \to TM. $$

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  • 1
    $\begingroup$ @Peter: I thought about this, but the problem is that one only has the cometric $g\in C^\infty(S^2(TM))$ as data. Of course, one can regard $g$ as a function on $T^\ast M$ that is quadratic on the fibers, but the Poisson bracket of $g$ with itself will vanish no matter what $g$ is. Moreover, because there are no first-order invariants if $g$ has full rank (by the Fundamental Lemma of Riemannian geometry), it's hard to see how one can avoid a construction of a first-order invariant that could detect the desired integrability without making an assumption of constant rank. $\endgroup$ – Robert Bryant Dec 29 '12 at 18:34
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    $\begingroup$ @Robert. Oh yes, in the skew case the grading is just what one needs, My mistake! Thanks. $\endgroup$ – Peter Michor Dec 29 '12 at 19:06
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    $\begingroup$ Thank you Robert, thank you Peter. I also started with Peter's idea precisely from that paper and got stuck with the fact that in the symmetric case one always gets zero from the bracket. However the second proposal with the two curvatures for the two distributions seems really conclusive. In the general case I was interested in perturbing a genuine Poisson structure with a symmetric tensor such that the integrability of the annihilator of the kernel is preserved. That would be a antisymmetric-symmetric version of bihamiltonian systems with hamiltonians given by the perturbed casimirs. $\endgroup$ – issoroloap Dec 30 '12 at 14:50
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    $\begingroup$ @issoroloap: You're welcome. Actually, I realized that my construction of an operator doesn't need symmetry or skew-symmetry either, and it doesn't require choosing any metric $g$, so I'll add that to my answer above. It gives an alternative to Peter's construction. Of course, his obstruction tensor and mine are `equivalent' in some sense, but the transformation between them will need to use the extra choices that Peter makes in his construction. $\endgroup$ – Robert Bryant Dec 30 '12 at 15:25
  • $\begingroup$ I was also wandering why you needed symmetry. I'll wait for your upgrde then! Thank you! $\endgroup$ – issoroloap Dec 30 '12 at 15:37

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 $Image(g^\sharp)\subset TM$?

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