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Question 1

In calculus, we visualize the tangent space $T_p(\mathbb R^n)$ at $p$ in $\mathbb R^n$ as the vector space of all arrows emanating from $p$. What is the motivation of creating of $T^*_p(\mathbb R^n)?$ How can we visualize covectors?

Question 2 How do I prove the sets $T^*_p(\mathbb R^n)$ are all disjoint? where $p\in U$

Considering $p,q\in U$ and $p\neq q$, suppose f is a non zero linear functional and $f \in T^*_p(\mathbb R^n)\cap T^*_q(\mathbb R^n)\implies f\in T^*_p(\mathbb R^n)$ and $f\in T^*_p(\mathbb R^n)$.How do I proceed further?. Can you please help me?

  • $\begingroup$ they are just linear functionals $T_p{\Bbb R^n}\to{\Bbb R}$ $\endgroup$
    – janmarqz
    Jul 7, 2018 at 15:30
  • $\begingroup$ What about the geometrical meaning? $\endgroup$
    – user464147
    Jul 7, 2018 at 15:32
  • $\begingroup$ They are employed to decompose the domain into level sets, which are lines $\endgroup$
    – janmarqz
    Jul 7, 2018 at 15:33
  • $\begingroup$ Can you explain? How is it coming? I am not able to visualize :( $\endgroup$
    – user464147
    Jul 7, 2018 at 15:39
  • $\begingroup$ my previous comment is a little wrong in the sense that I was thinking level set for $n=2$. For dimension $3$ the level sets are plane, for dimension $4$ hyperplanes an so on... In a while I will widen my comment into an answer. $\endgroup$
    – janmarqz
    Jul 7, 2018 at 15:55

3 Answers 3


I only given an answer to the first question. My answer is an addition to the answer given by janmarqz. Basically I just provide some plots for his second point.


I like the visualization given in V. Arnolds, Mathematical Methods of Classical Mechanics.

Suppose we have a covector $\omega \in T^*_0\mathbb{R}^2$, we then draw the hyperplanes $\dots, \omega^{-1}(\{-1\}),\omega^{-1}(\{0\}),\omega^{-1}(\{1\}), \omega^{-1}(\{2\}), \dots$ (Let's call them the contour lines of $\omega$.)

If we now want to see how the covector acts on a vector $\mathbf{v} \in T_0\mathbb{R}^2$, we just draw the arrow representing $\mathbf v$ and look how many contour lines we hit.

Below I have drawn this method for a basis $\{\omega, \eta\}$ of $T^*_0\mathbb{R}^2$.

Covectors in two dimensions

Why don't we draw arrows for covectors?

A typical difference between tangent vectors and cotangent vectors is their transformation behavior!

Let's use the linear transformation $\Phi:\mathbb R^2 \to \mathbb R^2: (x,y)^T \mapsto (x,2 y)^T$.

Vectors transform via the differential $T\Phi: T \mathbb R^2 \to T \mathbb R^2$ (sometimes also denoted by $D\Phi$). In our case this is just the Jacobian of $\Phi$ i.e. $\mathbf v \mapsto \begin{pmatrix}1&0\\0 & 2\end{pmatrix} \mathbf v$.

Covectors transform differently! An example is the gradient of a smooth function $f:\mathbb R^2 \to\mathbb R$. In classical calculus, we have see, that $\nabla f$ is always orthogonal to the contour-lines of $f$! But if you apply the transformation $\Phi$ and transform $\nabla f$ like a vector, it will not be orthogonal anymore!

enter image description here

But, if we visualise covectors by their contour-lines, they behave correctly! For example, if we draw the contour-lines of $df$, they will stay tangent to the contour-lines of $f$ after applying a transformation to the visualization, which is the correct behavior!

This is one of the reasons, why we treat vectors and covectors differently. And also the reason why we draw contour-lines of covectors instead of arrows.

Higher Dimensions

The reasoning above is still valid in higher dimensions, here the contour-lines become hyperplanes instead. (The example with the gradient is only plausible for Riemannian manifolds, but the example should just give a feeling how covectors are different than vectors. In general covectors transform according to the inverse of the adjoint of $T\Phi$, i.e. $\omega \mapsto ((T\Phi)^*)^{-1}$. )

  • $\begingroup$ What is $M$ here? $\endgroup$
    – user464147
    Jul 8, 2018 at 17:39
  • $\begingroup$ +1 excellent blowup! $\endgroup$
    – janmarqz
    Jul 8, 2018 at 18:56
  • 1
    $\begingroup$ $M = \mathbb R^2$, which plays the role of the manifold in this example. Note, that in this special case the manifold $\mathbb R^2$ and the 'attached' vector space $T_0 \mathbb R^2$ can be identified (also in a exact way), which allows use to draw contour lines of $f$ and contour lines of $df$ on top of each other. $\endgroup$ Jul 8, 2018 at 20:00
  • $\begingroup$ @janmarqz I'm glad that you like it. Thanks! $\endgroup$ Jul 8, 2018 at 20:02

Question 2 is easy: We already consider the tangent spaces $T_p{\mathbb R}^n$ as disjoint: Each point $p$ has its own tangent space. Of course all these tangent spaces are carbon copies of one and the same "model space" ${\mathbb R}^n$. (If we would not make the spaces $T_p{\mathbb R}^n$ disjoint a vector field would become a hedgehog!) Now each $T_p{\mathbb R}^n$ has its own dual space $\bigl(T_p{\mathbb R}^n\bigr)^*$, and these dual spaces are considered to be disjoint as well. This is matter of definition; no proof is needed.

Question 1 is more difficult. Number 1 there is no universally agreed way to "draw" a cotangent vector in a picture showing the vector space $V$, e.g., $T_p{\mathbb R}^n$. Sometimes you have a scalar product in $V$. It is then possible to identify each $\phi\in V^*$ with a vector $a_\phi\in V$, such that the identity $$\phi(x)=a_\phi\cdot x\qquad\forall\>x\in V$$ holds. E.g., if $\phi:=df(p)$ is the differential of a scalar function $f$ at $p$ then $a_\phi=\nabla f(p)$.

If you want an "intuitive" description of covectors $F\in V^*$ think of physical forces. Physicists cannot explain what a force is, for the same reasons that you have no intuitive feeling for covectors. They then draw the vector $a_F\in V$ and call it the force vector. In reality an (e.g., gravitational) force acting at $0\in V$ is a covector $F\in V^*$. If you want to move along the segment $\sigma$ connecting $0\in V$ with the point $x\in V$ then you have to perform the work $W_\sigma=F.x$ (or get back $W_\sigma$, depending on the sign conventions), whereby the $.$ denotes evaluation.

  • $\begingroup$ could you give your opinion about using level sets to attach some geometry to a 1-form, a covector at each position? $\endgroup$
    – janmarqz
    Jul 7, 2018 at 20:43
  • $\begingroup$ Are we considering $(T_p\mathbb R^n)^*$ disjoint or Can we able to rigorously prove it? If we consider functions from $(T_p\mathbb R^n)^*$ and $(T_q\mathbb R^n)^*$. the domain of consideration is different. hence, No two functions are equal. Am I correct? $\endgroup$
    – user464147
    Jul 8, 2018 at 2:32
  • $\begingroup$ We can rigorously prove it, your proof is fine; and it only depends on the $T_p(\mathbb{R}^n)$ being pairwise disjoint; and for this the proof depends on your initial definition of the tangent space $\endgroup$ Jul 8, 2018 at 9:31
  • $\begingroup$ Once we've decided that the tangent spaces are disjoint, we can prove that the cotangent spaces are disjoint. More generally, if two vector spaces $V$ and $W$ are disjoint (or merely distinct), then their duals $V^*$ and $W^*$ are disjoint also. $V^*$ consists of certain functions with domain $V$, and similarly for $W^*$. No function can have two different domains. $\endgroup$ Jul 8, 2018 at 13:17

There are few steps to consider:

1) How are the level sets of a linear maps $\Bbb R^2\to\Bbb R$?

2) How are the level sets of a linear maps $\Bbb R^3\to\Bbb R$?

3) How are the level sets of a linear maps $\Bbb R^n\to\Bbb R$?

4) The tangent spaces $T_p\Bbb R^n$ are vector spaces associated to a position $p$.

5) Different points in $\Bbb R^n$ receive different tangent spaces so they are disjoint.

6) The dual of a tangent space $T^*_p{\Bbb R^n}$ are linear maps $T_p{\Bbb R^n}\to\Bbb R$, so a 1-form is an association of a linear map to a copy of $\Bbb R^n$ attached to $p$ in which one gets the picture explained in points 1), 2) or 3) above.

What about an example?

Let's illustrate with a simple one-form $\omega=(5+x^2-y)dx+(9+x+y^2)dy$. With it you can prescribe to a points in $\Bbb R^2$ a linear functional, let's pick at the origin $\omega|_{0,0}=5dx+9dy$ which is the same as the linear map in $\Bbb R^2\to\Bbb R$ given by $$(x,y)\mapsto\omega|_{0,0}(x,y)=5x+9y$$ Here we can experiment a see how a little horizontal variation, say at $(\epsilon,0)$ what covector gonna be? This will be $$\omega|_{\varepsilon,0}(x,y)=(5+\varepsilon^2)x+(9+\varepsilon)y$$ which is $\approx5x+9y$, for $0<\varepsilon<\!\!<1$.


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