Why Is the Dual Basis Mathematically Unavoidable?

If we already have a basis for a vector space and we know how to express vectors in coordinates, why must we introduce an entirely separate space of covectors with its own basis?

You have an interest in differential geometry. A tangent space at a point on the manifold has no natural basis, so why do you want to single out just one choice of basis for it? You don't "already have" a basis for such vector spaces.

On finite-dimensional spaces, the nonzero elements of the dual space are all the possible coordinate functions for some basis (as the basis is allowed to vary). Confusing a vector space and its dual space is like confusing the role of $(a,b,c)$ and $(x,y,z)$ in equations of planes $ax + by + cz = d$. They really do serve different purposes.

What actually goes wrong if we refuse to distinguish vectors from covectors?

Concerning the distinction between vector spaces and dual spaces, the tangent and cotangent bundles of a manifold are built from tangent spaces and their dual spaces, respectively, and serve very different purposes: the tangent bundle is the setting for vector fields and the cotangent bundle (and its exterior powers) is the setting for differential forms.

One answer is implicitly already in the question, specifically in "If we already have a basis for a vector space and we know how to express vectors in coordinates, why must we introduce an entirely separate space of covectors with its own basis?" So you have a basis, say $\{e_1,\dots, e_n\}$ and you can express an arbitrary vector $x$ as a linear combination $\sum_{i=1}^n c_ie_i$. Of course, the coefficients $c_i$ here depend on $x$, so let me make that explicit by writing $c_i=f_i(x)$. Here each of the $n$ functions $f_i$ maps your vector space into the field of scalars, and it's easy to check that these functions are linear. At this point, you've arrived at the dual basis $\{f_1,\dots, f_n\}$. So this dual basis is implicit in the concepts of "basis" and "express in terms of coordinates"; the coordinates constitute the dual basis.

Why do we need a whole dual space, rather than just these particular covectors $f_i$ that constitute our dual basis? Because it's often convenient to change from a basis $\{e_1,\dots, e_n\}$ to some other basis, and that will lead to a new dual basis. All the covectors needed for such other bases constitute the whole dual space (except for the zero covector).

Sobolev spaces might answer this question for you: Essentially, if you consider continuous functions on a compact interval, then things like integrals of these functions against a measure are a linear functional and thereby part of the dual space. Continuous functions themselves are also in the dual, as you can integrate two continuous functions together on the same interval. However you cannot integrate two measures against each other. So the dual space is larger.

Sobolev spaces are function classes with varying levels of smoothness. The more smooth they are the larger the dual space. Whearas less smooth functions or even distributions/measures have smaller duals spaces.

• 1
  $\begingroup$ Yes, though "continuous functions themselves are also in the dual" is a bit problematic - technically speaking, there's just an embedding from $\mathcal{C}^1$ to $(H^1)^*$. Arguably it's ok to gloss over this, in the same way it's ok to gloss over the distinction between the natural number 3ドル,ドル the rational $\frac{3}1$ and the real 3ドル.000...$ - In the given example it's more dangerous though, because linear functionals and continuous functions are both functions, but with different types of arguments, which is quite confusing. $\endgroup$

  leftaroundabout

  – leftaroundabout

  2025年11月20日 10:04:20 +00:00

  Commented Nov 20 at 10:04
• $\begingroup$ Yes - there is also the inaccuracy that in the case $p=2$ the space is a hilbert space and thereby self-dual due to the Riez representation theorem. So in this sense you do not have a dual that is increasing in size as the function space becomes smoother. If I remember correctly, the insidious thing that is happening is that the linear functional generated by measures collapses together with linear functionals generated by functions when restricted to these hilbert spaces. In either case these spaces explain why it is important to make the distinction between the dual and the embedding. $\endgroup$

  Felix Benning

  – Felix Benning

  2025年11月20日 14:49:41 +00:00

  Commented Nov 20 at 14:49

In differential geometry, a vector bundle can be truly different (non-isomorphic) to its dual, even in finite dimension.

Take an example from complex geometry: the tautological line bundle of complex projective space $\mathbb{CP}^1$ admits no holomorphic global sections other than the zero section, whereas its dual has many.

More specifically, if you try to extend a local basis vector of the tautological line bundle holomorphically to the entire projective space, you will get a pole at some point. By contrast, if you try to do the same thing with a dual basis vector, you will obtain a zero at some point.

In fact, isomorphism classes of holomorphic line bundles form a group, called the Picard group, where dualizing the line bundle corresponds to taking the inverse.

What goes wrong if you refuse to "distinguish vectors from covectors" seems odd. The cheeky answer is that elements of the tangent space at a point of a smooth manifold and elements of the cotangent space at a point of a smooth manifold belong to different vector spaces, so you're not recognizing the actual vector space structure on those sets. But another answer is this: I have faith that you understand a smooth vector field on a smooth manifold is defined to be a smooth section of the tangent bundle. Well, a smooth covector field on a smooth manifold is defined to be a smooth section of the cotangent bundle. Smooth covector fields pop up enough in application within sciences that it's merited an independent study of "smooth rank $k$ subbundles of the cotangent bundle". The cotangent bundle inevitably starts off with the cotangent space. In finite-dimensions, if you give me a basis for the tangent space at a point of a finite-dimensional smooth manifold then you can construct a basis called a dual basis for its dual vector space... being the cotangent space. Sometimes working with this dual basis can be useful to deduce additional properties. In short, you would be missing out on mathematical objects which are used to model real world phenomenon in sciences and engineering. This assumes you're interested in actually focusing your differential geometry/topology knowledge to application. It is likely worthwhile to take a science or engineering class or an applied mathematics class with intent on exposing you to mathematics as it pertains to science/engineering if that's your goal. Otherwise, you could just go ahead and study vector bundles or smooth subbundles of the cotangent bundle or geometry pertaining to the canonical symplectic structure on the cotangent bundle. Technically you could even study the geometry of smooth Riemannian metrics on the cotangent bundle to a smooth finite-dimensional manifold... people usually put a smooth Riemannian metric on the tangent bundle rather than the cotangent bundle. Existence of a smooth Riemannian metric à la Spivak.

• $\begingroup$ Not every applied mathematics class actually... focuses on applications. Sometimes "applied mathematics" is an excuse to tuck away all of the differential equation classes and force the "pure" mathematics people to slave away at cohomology and $p$-adic structures. Some applied mathematics classes are literally just about the theory of differential equations and dynamical systems. So, when I say take a class in applied mathematics with intent on exposing you to mathematics as it pertains to science/engineering I mean take one that actually makes you integrate science/engineering into the math. $\endgroup$

  Man-I-Fold

  – Man-I-Fold

  2025年11月23日 13:31:44 +00:00

  Commented 19 hours ago

• linear-algebra
• differential-geometry