For 1-forms, you can get some intuition for exterior differentiation from how it shows up in Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:

Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.

Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form a foliation by hypersurfaces in $M$. For a hyperplane field defined by a single 1-form one can fit the planes together if and only if $d\alpha$ mod $\alpha$ is zero. This is usually expressed by the vanishing of $\alpha\wedge d\alpha$.

(In the general case, where instead of $\alpha$ we have a set of linearly independent 1-forms $\{\alpha_j\}_{j=1}^r$, the ideal in the algebra of differential forms on $M$ generated by $\{\alpha_j\}_{j=1}^r$ must be closed under exterior differentiation; equivalently $d\alpha_j\wedge\alpha_1\wedge\cdots\wedge\alpha_r=0$ for all $j$).

Two simple examples:

(1) if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const (and of course $d\alpha=0$).

(2) If $\alpha = g df$ for some non-vanishing function $g$, e.g. $\alpha=ydx$ in the upper half plane of $\mathbb{R}^2$, then this is just as good, since ker $\alpha$ is still tangent to $f=$const. Note that $d\alpha=dg\wedge df=(dg/g)\wedge\alpha$, which vanishes mod $\alpha$ and thus $\alpha\wedge d\alpha=0$.

Hence $\alpha\wedge d\alpha$, or $d\alpha$ mod $\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.

(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners. Thanks to Marcos Cossarini and Ben McKay for pointing out in the comments that the original version of this was wrong!)

Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $\alpha\wedge d\alpha = dz\wedge dx \wedge dy$:

standard contact structure on R^3 http://upload.wikimedia.org/wikipedia/en/thumb/f/f5/Standard_contact_structure.svg/500px-Standard_contact_structure.svg.png

For 1-forms, you can get some intuition for exterior differentiation from how it shows up in Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:

Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.

Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form a foliation by hypersurfaces in $M$. For a hyperplane field defined by a single 1-form one can fit the planes together if and only if $d\alpha$ mod $\alpha$ is zero. This is usually expressed by the vanishing of $\alpha\wedge d\alpha$.

(In the general case, where instead of $\alpha$ we have a set of linearly independent 1-forms $\{\alpha_j\}_{j=1}^r$, the ideal in the algebra of differential forms on $M$ generated by $\{\alpha_j\}_{j=1}^r$ must be closed under exterior differentiation; equivalently $d\alpha_j\wedge\alpha_1\wedge\cdots\wedge\alpha_r=0$ for all $j$).

Two simple examples:

(1) if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const (and of course $d\alpha=0$).

(2) If $\alpha = g df$ for some non-vanishing function $g$, e.g. $\alpha=ydx$ in the upper half plane of $\mathbb{R}^2$, then this is just as good, since ker $\alpha$ is still tangent to $f=$const. Note that $d\alpha=dg\wedge df=(dg/g)\wedge\alpha$, which vanishes mod $\alpha$ and thus $\alpha\wedge d\alpha=0$.

Hence $\alpha\wedge d\alpha$, or $d\alpha$ mod $\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.

(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners. Thanks to Marcos Cossarini and Ben McKay for pointing out in the comments that the original version of this was wrong!)

Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $\alpha\wedge d\alpha = dz\wedge dx \wedge dy$:

For 1-forms, you can get some intuition from Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:

Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.

Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form hypersurfaces in $M$. In this case, it turns out that one can fit the planes together if and only if $d\alpha\equiv0$.  (In the general case, where instead of $\alpha$ we have an algebraic ideal of 1-forms $\mathcal I$, this is $d\mathcal I\equiv 0\mod\mathcal I$).

Here's some more discussion: if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const. Similarly, if $d\alpha=0$ then it's clear that we can find such $f$ locally (not globally if $\alpha$ isn't exact).

Hence $d\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.

(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners).

Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $d\alpha = dx dy$:

standard contact structure on R^3 http://upload.wikimedia.org/wikipedia/en/thumb/f/f5/Standard_contact_structure.svg/500px-Standard_contact_structure.svg.png

For 1-forms, you can get some intuition for exterior differentiation from how it shows up in Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:

Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.

Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form a foliation by hypersurfaces in $M$. For a hyperplane field defined by a single 1-form one can fit the planes together if and only if $d\alpha$ mod $\alpha$ is zero. This is usually expressed by the vanishing of $\alpha\wedge d\alpha$.

(In the general case, where instead of $\alpha$ we have a set of linearly independent 1-forms $\{\alpha_j\}_{j=1}^r$, the ideal in the algebra of differential forms on $M$ generated by $\{\alpha_j\}_{j=1}^r$ must be closed under exterior differentiation; equivalently $d\alpha_j\wedge\alpha_1\wedge\cdots\wedge\alpha_r=0$ for all $j$).

Two simple examples:

(1) if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const (and of course $d\alpha=0$).

(2) If $\alpha = g df$ for some non-vanishing function $g$, e.g. $\alpha=ydx$ in the upper half plane of $\mathbb{R}^2$, then this is just as good, since ker $\alpha$ is still tangent to $f=$const. Note that $d\alpha=dg\wedge df=(dg/g)\wedge\alpha$, which vanishes mod $\alpha$ and thus $\alpha\wedge d\alpha=0$.

Hence $\alpha\wedge d\alpha$, or $d\alpha$ mod $\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.

(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners. Thanks to Marcos Cossarini and Ben McKay for pointing out in the comments that the original version of this was wrong!)

Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $\alpha\wedge d\alpha = dz\wedge dx \wedge dy$:

standard contact structure on R^3 http://upload.wikimedia.org/wikipedia/en/thumb/f/f5/Standard_contact_structure.svg/500px-Standard_contact_structure.svg.png