# What is an ordered basis?

To my understanding, $$e_1 = \{1,0,\ldots,0\},\quad e_2 = \{0,1,\ldots,0\}, \quad \ldots, \quad e_n = \{0,0,\ldots,1\}$$ is an ordered basis for a vector space of dimension $n$. But the group of basis, which implies no ordering is not ordered.

For example: $$\{1,0,\ldots,0\},\quad \{0,1,\ldots,0\},\quad \ldots, \quad\{0,0,\ldots,1\}$$ is not ordered basis, but a basis, for a vector space of dimension $n$.

Is this correct? Thanks!

The difference, of course, is the ordering. An ordered basis $B$ of a vector space $V$ is a basis of $V$ where some extra information is provided: namely, which element of $B$ comes "first", which comes "second", etc. If $V$ is finite-dimensional, one approach would be to make $B$ an ordered $n$-tuple, or more generally, we could provide a total order on $B$.

The difference can be obscured however by common abuse of notation and abuse of terminology regarding bases vs. ordered bases. For example, it would be quite common to say that $$\{e_1,\ldots,e_n\}$$ is an ordered basis, even though it is just a set and not an ordered $n$-tuple, because the indexes tell you what the intended ordering is.

Let's consider $\mathbb{R}^3$. An example of a plain-old, vanilla basis would be a set $$\left\{\begin{array}{ccc} & a & \\ & & \alpha\\ \mathbf{a}& & \end{array}\right\}$$ where $a$, $\alpha$, and $\mathbf{a}$ are a basis of $\mathbb{R}^3$ (which I have intentionally written and labeled in a way such that there would be no implied order). An example of an ordered basis could be $$(a,b,c)$$ which is an ordered $3$-tuple where $a$ comes first, $b$ comes second, and $c$ comes third (reinforced by the fact that that is in alphabetical order), or equivalently, I could have said $$\{a,b,c\}$$ is an ordered basis where the ordering is specified by $a<b<c$.

• Could you give an example of what is not an ordered basis? Thanks Zev. – 1LiterTears Jun 18 '13 at 19:11
• @MathSnail $\left\{(0, 1), (1, 0)\right\}$ – Jack M Jun 18 '13 at 19:13

You need a one to one correspondence with $\mathbb{N}$ for it to be ordered. The second example can be ordered but as how it stands it isn't quite an ordered basis just yet.

As stated above, one way to define an ordered basis would be a basis $B$ together with a total order on $B$.

Let us suppose that $V$ is a vector space over a field $K$ (which could be $\mathbb{R}$ or $\mathbb{C}$ for example). If $V = K^n$, we automatically have an ordered basis of $V$ (in the above sense), namely the standard basis: $$\vec{e_1} = (1,0,\dots,0) \quad \vec{e_2} = (0,1,\dots,0) \quad \dots \quad \vec{e_n} = (0,\dots,0,1)\\ B = \{ \vec{e_1}, \dots, \vec{e_n} \},$$ with the implied ordering: $\vec{e_i} \leq \vec{e_j} \iff i \leq j$.

Our next observation will be that if you have two vector spaces $V_1, V_2$ over $K$, a basis $B_1$ of $V_1$, and an isomorphism $\Phi: V_1 \to V_2$, then $B_2 = \Phi(V_1) \subseteq V_2$ will be a basis for $V_2$. Also, if $B_1$ is ordered, we can define an order on $B_2$ by pulling the basis vectors of $B_2$ back through $\Phi^{-1}$.

Hence, if our vector space $V$ is finite-dimensional ($\dim(V) = n$), given an ordered basis $B$ we can construct an isomorphism $\Phi: K^n \to V$ sending the standard basis to $B$, which is order-preserving on the standard basis. Conversely, any isomorphism $\Phi: K^n \to V$ induces an ordered basis by letting $B = \Phi(\{\vec{e_1},\dots,\vec{e_n}\})$ and using the pullback-order. In this sense, an ordered basis of $V$ is equivalent to an isomorphism $\Phi: K^n \to V$.

Another way of looking at this can be obtained by observing that any two isomorphisms $\Phi_1, \Phi_2: W_1 \to W_2$ of vector spaces $W_1$ and $W_2$ are related by an isomorphism $\Psi = \Phi_2 \circ \Phi_1^{-1} : W_2 \to W_2$, so that $\Phi_2 = \Psi \circ \Phi_1$. So, if you have defined an ordered basis on $V$ by picking an isomorphism $\Phi: K^n \to V$, any other ordered basis can be obtained by specifying an isomorphism $\Psi \in GL(V)$ (the group of linear automorphisms / general linear group on $V$). Hence, an ordered basis can be abstracted to mean any particular automorphism of $V$ together with some "standard" basis of $V$.

The isomorphism approach is particularly useful in differential geometry, to define the frame bundle on a manifold. This makes it possible to define a frame at each point and gluing them together as a smooth fiber bundle without worrying about the differentiability of some total order.

A basis is a set of vectors that spans a vector space (or vector subspace), each vector inside can be written as a linear combination of the basis, the scalars multiplying each vector in the linear combination are known as the coordinates of the written vector; if the order of vectors is changed in the basis, then the coordinates needs to be changed accordingly in the new order.

That is the reason why the authors say: "an ordered basis", when they write about, by example, a basis $\{v_1, v_2, \dots, v_n\}$ then it is implicit that an order is said and it is an ordered basis.