(The approach used on this page is similar to the approaches used by [MTW
Fundamentally, a 1-form can be imagined to be a gradient.
Cover the manifold
with numbers. That's a "scalar field". The gradient is the
rate at which those numbers are changing at each point.
Let our "scalar field" represent height
. Then the
"gradient" of the scalar field is exactly what we think of as a
gradient in "real life". When we look at a point on a hillside,
the "gradient" is a measure of which way the hill slopes "up", and how
steep the slope is.
The most natural way to represent a gradient is as a series of contour
This is how gradients have been drawn on maps of the
Earth for centuries. In more than two dimensions, the "contour
lines" are actually hypersurfaces, but the principle is unchanged.
Locally, the best linear approximation of the contours is a collection
of straight lines (or hyperplanes). At a particular point on the
manifold, the local linear approximation to the contours is a
"covector" (or dual vector) and it's represented as a collection of
hyperplanes in the tangent space at that point. On this page, all
drawings are restricted to 2 dimensions, and the hyperplanes appear as
Now, having given this description, I must hasten to add that, while
every gradient is a 1-form, not
all 1-forms are actually the gradients of functions. In
only "closed", or curl-free, 1-forms are actually the gradients of
functions. But for the purposes of understanding 1-forms and
action on vectors, it's perfectly reasonable to think in terms of
gradients. And the covector associated with a 1-form at each
point is the same, whether or not the 1-form globally represents a
The gist of this page is in the pictures.
Definition of a Covector
A "covector" is a rank (0,1) tensor,
also called a "dual vector".
It is a linear function
which maps vectors into real numbers. The term "dual vector" is
actually more popular than "covector" but seems cumbersome to me.
"Covector" also suggests "covariant vector", in contrast to ordinary
vectors, which are "contravariant".
A covector provides a way to measure
a vector. If
we visualize it, as Misner, Thorne, and Wheeler do, as a collection of
parallel hyperplanes (see figure 1)
, then the length of a vector
corresponds to the
number of hyperplanes "pierced" by the vector.
Figure 1 -- A covector:
Alternatively, we can view it as a "spatial frequency". It's a
a direction. The length of a vector is then
the number of wavecrests the vector crosses.
In Figure 2 we show a the covector being "applied" to a
vector. The result is the number of hyperplanes the vector
crosses. In the figure, the product of the covector and the
vector is 5. In other words, the length of the vector along
the direction of the covector
is 5 times the "wavelength" of the
It's worth pointing out that, if we were to multiply the covector by
2, we would double
the number of blue lines -- the distance
between the lines would decrease
if the magnitude of the
. Once again, the covector is a
spatial frequency; double it, and the density of wavecrests increases;
the "wavelength", on the other hand, decreases when we do that.
Figure 2 -- A Covector Applied to a Vector:
Taking the analogy one step farther, a covector is a
frequency, a vector is like a wavelength, and their "product" is
analagous to a velocity.
So far we've talked about covectors with no mention of
coordinates. In a particular coordinate system, the
"coordinates" of a covector are coefficients on the basis covectors,
just as the "coordinates" of a vector are coefficients on the
basis vectors. The "basis covector" corresponding to a
particular "basis vector" is the covector which points in the same
direction as that basis vector, and which has a wavelength exactly as
long as the basis vector. No mention of a "metric" is made here;
the metric doesn't enter into it. (Under rigid rotations, the
correspondence between basis vectors and basis covectors is
fixed. However, if we allow more general transformations, then
is dependent on the chosen
basis. In the absence of a metric, there is no distinguished
isomorphism between covectors and vectors.)
This may sound confusing, but hopefully a picture will make it clearer.
Figure 3 -- A Covector, with basis vectors:
To find the coordinates of a particular covector, all we need to do
is apply it to each of the basis vectors
. The length of
each basis vector, as measured by the covector, is the coefficient
of the corresponding basis covector. In Figure 3, we show the
coordinates of our example covector in terms of a particular
coordinate system. Basis vector ex
crosses 2 lines,
and basis vector ey
crosses about 2.3 lines. So, the
components of the covector, in this coordinate system,
A "1-form" is a covector field. In other words, a 1-form
associates a covector with each point on the manifold
-- or each
point in spacetime, if you prefer. Most of the time, no
distinction is drawn between a 1-form and the covector associated
with it at a particular point, and the term "covector" is, in fact,
hardly ever used in some texts.
Throughout this website, I tend to use "1-form" to refer to both the
covector field on the manifold, and the covector associated with that
field at each point.
Dot Product and Rigid Rotations
Figure 4 -- A Covector on a Vector, With Coordinates
The dot product of a vector and a covector, or the result of
applying a 1-form to a vector, is just the number of lines of the
1-form which the vector crosses. It's the length
vector, as measured by the 1-form.
If we look at the 1-form and the vector in a particular coordinate
system, then we see that the total number of lines the vector crosses
is just the number of lines it crosses in the direction of each basis
vector in turn. It doesn't matter what path you follow -- you
need to cross the same number of lines no matter how you go! The
number of lines crossed in the direction of each basis vector is just
the number of lines crossed by that basis vector, times
distance the vector travels in along that axis. In other words,
it's the product of the corresponding component of the vector, with the
corresponding component of the covector.
Or, again in other words, it's just the dot product of the vector and
It doesn't matter how the axes are chosen.
In Cartesian space, under rigid rotations, the basis covectors
are naturally isomorphic with the basis vectors,
vector has a natural covector corresponding to it, which has the
same components as the vector. In this case, we have just
exhibited a simple visual proof that dot product is unaffected by rigid