Basis Vector Length: Unit Length & Mistakes to Avoid

[snoopies622]

For an ordinary vector V, the square of its length is $$V \cdot V = V^a V_a$$.

For basis vectors, $$e^a \cdot e_b = \delta ^a _b$$ so $$e^a \cdot e_a = 1$$.

Since $$1^2 = 1$$, this implies that every basis vector is of unit length.

What is my mistake?

 

[waht]

it's fine

 

[atyy]

For an ordinary vector V, the square of its length is $$V \cdot V = V^a V_a$$.

For basis vectors, $$e^a \cdot e_b = \delta ^a _b$$ so $$e^a \cdot e_a = 1$$.

Since $$1^2 = 1$$, this implies that every basis vector is of unit length.

What is my mistake?

Using the formula on top, the length of a basis vector is $$e_a \cdot e_a$$

The equation for unit vectors perpendicular to each other is $$e_a \cdot e_b = \delta _a _b$$

The equation for non-unit non-perpendicular vectors is e_a.e_b= g_ab

Given a set of basis vectors, the dual basis covectors are defined by $$e^a \cdot e_b = \delta ^a _b$$

Or something like that, I can never quite remember which indices go up or down.

 

[HallsofIvy]

For an ordinary vector V, the square of its length is $$V \cdot V = V^a V_a$$.

For basis vectors, $$e^a \cdot e_b = \delta ^a _b$$ so $$e^a \cdot e_a = 1$$.

Since $$1^2 = 1$$, this implies that every basis vector is of unit length.

What is my mistake?

In order to say $$e^a \cdot e_b = \delta ^a _b$$ or $$e^a \cdot e_a = 1$$ you have to assume an orthonormal basis. Essentially what you are saying is "Assuming basis vectors have unit length, then every basis vector is of unit length"!

 

[snoopies622]

In order to say $$e^a \cdot e_b = \delta ^a _b$$ or $$e^a \cdot e_a = 1$$ you have to assume an orthonormal basis.

I did not know this. If my basis is not orthonormal and I have chosen my (say) countervariant basis vectors, how do I find their covariant counterparts? Does one raise or lower indices in the same way as with ordinary vectors/tensors, or is it different with basis vectors?

For example, suppose I have a two-dimensional manifold and a coordinate chart (u,v) with metric $$g_{uu}=2$$ $$g_{uv}=g_{vu}=-2$$ $$g_{vv}=4$$ and I choose $$e^u =<1,0>$$ $$e^v =<0,1>$$. How do I find $$e_u$$ and $$e_v$$?

 

[HallsofIvy]

I mispoke before because I did not realize that you were talking about a product of "vectors" and "co-vectors" or the dual space.

The dual space of a vector space, V, is the set of linear functions from V to its underlying field.

In fact, given any basis for a vector space, the the corresponding basis for the dual space is defined by "e_i(e^j)= 1" and then that is used to define the "dot product". However, that dot product depends upon the choice of basis! In that sense, yes, every basis vector has unit length: the choice of dot product based on that basis guarentees that.

If you have a given dot product (perhaps based on some basis) and use it with a different basis, It does not follow that $e^i\cdot e_j= \delta^i_j$

$e_i= g_{ij}e^j$. In the case you give, it is just a matrix multiplication:
$$e^u= \left[\begin{array}{cc}2 & -2 \\ -2 & 2\end{array}\right]\left[\begin{array}{c}1 \\ 0\end{array}\right]= <2, -2>$$
Similarly, $e^v$ is <-2, 2>.

 

[snoopies622]

Is the dual basis the same thing as the "reciprocal basis" described here?

http://home.pacbell.net/bbowen/covariant.htm

 

[atyy]

Is the dual basis the same thing as the "reciprocal basis" described here?

http://home.pacbell.net/bbowen/covariant.htm

Yes, the "reciprocal basis" is the same as a "dual basis". The reciprocal basis only exists when you have already defined one basis.

 

[snoopies622]

A clarification, please: According to the description here

http://home.pacbell.net/bbowen/covariant.htm

And the drawing here

http://en.wikipedia.org/wiki/Image:Basis.gif

when one raises or lowers the index on a vector/covector, one is still describing the same arrow, only using different ‘building block’ arrows to do so. But when one raises or lowers the index on a basis-vector/basis-covector, one is actually turning it into a different arrow – one that goes from being parallel to a coordinate line to one that is perpendicular to the other coordinate line(s) or vice versa.

Granted that vectors/co-vectors are not really arrows, is this otherwise correct?

 

[atyy]

But when one raises or lowers the index on a basis-vector/basis-covector, one is actually turning it into a different arrow

Using Bowen's notation and formulas, if V=A₁, its contravariant components are [V¹=1, V²=0]. To figure out its covariant components we lower its index:

A₁ = V¹A₁+V²A₂ = V₁A¹+V₂A² = Vⁱ*G_i1A¹+Vⁱ*G_i2A²
= (V¹*G₁₁+V²*G₂₁)A¹+(V¹*G₁₂+V²*G₂₂)A²
= G₁₁A¹+G₁₂A²

So if {A₁,A₂} are the basis vectors, their contravariant components are {[1,0], [0,1]}, and their covariant components are {[G₁₁,G₁₂]_c,[G₂₁,G₂₂]_c}, so that A_i.A_j=G_ij

The sets of basis vectors {A₁,A₂} and {A¹,A²} are reciprocal, but lowering the index on A₁ doesn't change it into A¹, it still remains itself described in terms of the reciprocal basis vectors, just as with any other vector.

The process of getting a set of reciprocal basis vectors given a set of basis vectors is a different process from raising or lowering an index.

(Actually, I usually think of vectors and covectors as being in different spaces. The vectors are column vectors and the covectors are row vectors. You can multiply a column and row vector to get a number without any metric. Without a metric, the column vectors and row vectors are not related, unless we define a basis for the column vectors and a reciprocal basis for the row vectors, which means the relationship between column vectors and row vectors changes with the basis. Without a metric you cannot multiply two column vectors to get a number. Once you have a metric, you can use that to multiply two column vectors to get a number, by using it to change one column vector into a row vector. You can also use the metric to enforce a fixed relationship between column vectors and row vectors, so you can identify a vector with its covector.)

 

[snoopies622]

The sets of basis vectors {A₁,A₂} and {A¹,A²} are reciprocal, but lowering the index on A₁ doesn't change it into A¹, it still remains itself described in terms of the reciprocal basis vectors, just as with any other vector.

The process of getting a set of reciprocal basis vectors given a set of basis vectors is a different process from raising or lowering an index.

So an expression like $$e^u = <1,0>$$ is ambiguous since the raised u could mean either that the components of the u basis vector (or covector) are contravariant, or that the basis vector itself is contravariant and the components are either contravariant or covariant?

(Edit: I guess "expression" should be "equation".)

 

[HallsofIvy]

Yes, that's true.

 

[snoopies622]

So does $$\frac {\partial}{\partial t}$$ represent the either contravariant or covariant components of a covariant basis vector, or the covariant components of a basis vector that could be either contravariant or covariant?

 

[atyy]

So does $$\frac {\partial}{\partial t}$$ represent the either contravariant or covariant components of a covariant basis vector, or the covariant components of a basis vector that could be either contravariant or covariant?

Given coordinates $$\{t,x\}$$ on a manifold, the basis vectors can be chosen to be $$\{e_0={\frac {\partial}{\partial t}},e_1={\frac {\partial}{\partial x}}\}$$, and they can be used to represent any vector using contravariant components: $$v=v^ie_i=v^i\frac {\partial}{\partial x_i}=v^i$$ , where in the final step I omitted the basis vectors only for notational convenience as is often done.

 

[snoopies622]

So an expression like $$e^u = <1,0>$$ is ambiguous since the raised u could mean either that the components of the u basis vector (or covector) are contravariant, or that the basis vector itself is contravariant and the components are either contravariant or covariant?

Wait, it's not ambiguous. If $$e^u = <1,0>$$ then it is being expressed in terms of the e^u and e^v basis vectors. e^u=(1)(e^u)+(0)(e^v). I may actually understanding this now.