# Different definitions of covector

In the book An introduction to manifolds by Tu, Loring W, covector is defined as follows:

If $$V$$ and $$W$$ are real vector spaces, we denote by $$\operatorname{Hom}(V, W)$$ the vector space of all linear maps $$f: V \rightarrow W$$. Define the dual space $$V^{\vee}$$ of $$V$$ to be the vector space of all real-valued linear functions on $$V$$ : $$V^{\vee}=\operatorname{Hom}(V, \mathbb{R})$$ The elements of $$V^{\vee}$$ are called covectors or $$1$$-covectors on $$V$$.

However, in the book Quick Introduction to Tensor Analysis by R. A. Sharipov, the covector is defined as follows:

Let's denote our hypothetical object by $$\mathbf{a},$$ and denote by $$a_{1}, a_{2}, a_{3}$$ that three numbers which represent this object in the basis $$\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$$. By analogy with vectors we shall call them coordinates. But in contrast to vectors, we intentionally used lower indices when denoting them by $$a_{1}, a_{2}, a_{3}$$. Let's prescribe the following transformation rules to $$a_{1}, a_{2}, a_{3}$$ when we change $$\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$$ to $$\tilde{\mathbf{e}}_{1}, \tilde{\mathbf{e}}_{2}, \tilde{\mathbf{e}}_{3}$$: $$\tilde{a}_{j} =\sum_{i=1}^{3} S_{j}^{i} a_{i} \tag{8.1}$$ $$a_{j} =\sum_{i=1}^{3} T_{j}^{i} \tilde{a}_{i}\tag{8.2}$$ Here $$S$$ and $$T$$ are the same transition matrices as in case of the vectors in $$(6.2)$$ and $$(6.5)$$. Note that $$(8.1)$$ is sufficient, formula $$(8.2)$$ is derived from $$(8.1)$$.

DEFINITION 8.1. A geometric object a in each basis represented by a triple of coordinates $$a_{1}, a_{2}, a_{3}$$ and such that its coordinates obey transformation rules $$(8.1)$$ and $$(8.2)$$ under a change of basis is called a covector.

My question is, are the two definitions equivalent? What's the relation between them? I cannot figure out the relation between "obeying transformation rules" and linear maps $$f: V \rightarrow W$$.

• the definition of Loring Tu is the general one, the other could be a particular case of covectors in $\Bbb R^3$ Commented Jan 16, 2021 at 9:52
• Could you expain concretely about how the other one be a particular case? Commented Jan 16, 2021 at 10:19
• Presumably those transformation rules give rise to the identity $a\cdot v=\tilde{a}\cdot\tilde{v}$ where $v$ and $\tilde{v}$ both refer to the same vector expressed in the different bases. Then the operation (dot product with $a$ or $\tilde{a}$) is independent of basis and can be regarded as a linear functional $V\to\mathbb{R}$ à la Tu.
– Ali
Commented Jan 16, 2021 at 11:20
• Ok I've deleted my addition notes. Commented Jan 16, 2021 at 15:17
• @FFjet maybe the notes refer to this. Then it seems that the covectors are a case of covariant vectors. Commented Jan 16, 2021 at 15:19

The idea is that $$V^\vee:=\operatorname{Hom}(V,\mathbf{R})$$ is a vector space, and thus vectors in $$V^\vee$$ can have components with respect to a basis. If we have two different bases for $$V^\vee$$, then we will get two different sets of components for the same vector, and the way these components are related to each other is what we call a transformation law, or a change of basis, or something along those lines.
Let's look at the vector space $$\mathbf{R}^3$$. If we have two bases $$E=\{e_1,e_2,e_3\}$$ and $$F=\{f_1,f_2,f_3\}$$, we may express some vector $$v\in\mathbf{R}^3$$ as $$v=a^1e_1+a^2e_2+a^3e_3$$ or $$b^1f_1+b^2f_2+b^3f_3$$. How are the $$a^i$$'s and $$b^i$$'s related? Well, it's enough to know how $$E$$ and $$F$$ are related; so let $$A$$ be the matrix sending $$e_i$$ to $$f_i$$. That is, $$f_i=Ae_i=A^1_ie_1+A^2_ie_2+A^3_ie_3$$. Let's write $$\epsilon^j$$ for the map that takes a vector to its $$j$$-th component in the basis $$E$$, so $$v=\epsilon^1(v)e_1+\epsilon^2(v)e_2+\epsilon^3(v)e_3$$. Notice that $$\epsilon^i$$ is linear, and $$\epsilon^j(Ae_i)=A^j_i$$. We then compute \begin{align*} a^j=\epsilon^j(b^1f_1+b^2f_2+b^3f_3) &=b^1\epsilon^j(f_1)+b^2\epsilon^j(f_2)+b^3\epsilon^j(f_3)\\ &=b^1\epsilon^j(Ae_1)+b^2\epsilon^j(Ae_2)+b^3\epsilon^j(Ae_3)\\ &=b^1A^j_1+b^2A^j_2+b^3A^j_3. \end{align*}
So $$a^j=\sum_ib^iA^j_i$$. Roughly speaking, if $$F$$ is the old basis system and $$E$$ is the new one, this formula tells us that we must take the matrix $$A$$ sending new basis vectors $$e_i$$ to old ones $$f_i$$ and multiply it with the old components $$b^i$$ to get the new components $$a^i$$. As such, we say that the components of a vector transform contravariantly. (Later you'll find that the components of a covector transform covariantly. And a vector transforms covariantly as well, except people sometimes will say vector when they mean the components of the vector. And so they'll say a vector transforms contravariantly.)
So we can think of a vector as an abstract element of the vector space $$\mathbf{R}^n$$ subject to the vector space axioms, or as an equivalence class of $$n$$-tuples as subject to the constraints of transformation laws. Both ways end up imposing the same structure. Similarly for your question on covectors being linear maps from $$V$$ to $$\mathbf{R}$$. You should be working on a higher level of abstraction and thinking of covectors as elements of vector spaces, and thus admitting transformation laws with respect to different bases.