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I am reading this paper Sigma Coordinate - Contravariance and covariance and I understand how covariant and contravariant vectors are defined mathematically Covariance and Contravariance and I had some questions regarding how the authors are using those definitions for testing orthogonality. The problem domain is in the Euclidean space and Cartesian axes. I have curvilinear coordinates and I can define a set of covariant and contravariant basis vectors and they form a set of mutually orthonormal basis vectors defined by the Kronecker delta.

Now one of the axes is no longer orthogonal to the other two and this is the sigma coordinate as defined in that paper.

$\sigma_z$ = H. z-h/H-h where H is model top and h = h(x,y)

Is it always the case when you have a non orthgonal coordinate surface the basis vectors that are orthogonal to each other will transform in a covariant manner and the non orthogonal basis vector will transform in a contravariant manner ?

If not can someone explain what this text means

In a $\sigma$-coordinate, the horizontal covariant basis vectors and the vertical contravariant basis vectors vary in the horizontal and vertical, respectively, while the covariant and contravariant basis vectors are non-orthogonal when the height and slope of terrain do not equal zero

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  • $\begingroup$ Aren't the basis vectors for both contravariant and covariant cases non orthogonal? $\endgroup$
    – Courage
    Commented Jan 31, 2016 at 2:58
  • $\begingroup$ @TheGhostOfPerdition - I am not sure. $\endgroup$
    – user66043
    Commented Jan 31, 2016 at 3:06

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"Contraction-orthogonality" of covariant and contravariant basis

Contravariant vectors or just "vectors" are defined as elements of the tangent space at a given point. In practice, they are defined with respect to a coordinate-vector basis $\mathbf{e}_{(i)}$, where $\mathbf{e}_{(i)}$ is the vector tangent to the $i$-th coordinate line. Then they are given, as per usual, as a linear combination of the basis vectors (Einstein summation assumed henceforth) $$\mathbf{v} = v^i \mathbf{e}_{(i)}$$

Covariant vectors or "1-forms", on the other hand, are more abstract objects which are defined exclusively through their action on contravariant vectors. For instance, a 1-form acting on a (contravariant vector) will return a number $$\mathbf{\alpha}(\mathbf{v}) = C,\,C \in \mathbb{R}$$

Another requirement on the 1-form $\alpha$ is that its action is linear, that is for any vectors $\mathbf{v},\,\mathbf{w}$ and any two constant $D$, $E$ $$\alpha(D \mathbf{v} + E \mathbf{w}) = D \alpha(\mathbf{v}) + E \alpha(\mathbf{w})$$ This actually means we can completely reconstruct its action from the components $\alpha_i \equiv \alpha(\mathbf{e}_{(i)})$ because then thanks to the linearity (verify this yourself) $$\alpha(\mathbf{v}) = \alpha_i v^i$$ Alternatively, you can define $\alpha_i$ as components with respect to a covariant basis $\alpha = \alpha_i \epsilon^{(i)}$ where $\epsilon^{(i)}$ is defined by the property $$\epsilon^{(i)}(\mathbf{e}_{(j)}) = \delta^i_j$$ That is, the "contraction-orthogonality" of covariant and contravariant bases has nothing to do with a geometrical definition of distance or angles (AKA the metric).


Raising and lowering indices

Now consider a space where we actually have a metric $g_{ij}$, abstractly we can define it as $$\mathbf{g}(\mathbf{v},\mathbf{w})=C,\,C \in \mathbb{R}$$ where we again require linearity. Now consider the form $\kappa$ defined by $\kappa(\mathbf{w}) \equiv \mathbf{g}(\mathbf{u},\mathbf{w})$ for some vector $\mathbf{u}$. $\kappa$ is a full-fledged form as defined above, its components can be shown to be $$\kappa_i = g_{ij} u^j$$ physicist know this as "index lowering". We assume that the metric is non-degenerate which means that $u \leftrightarrow \kappa$ is a one-to-one relation and we can find an inverse $\mathbf{g}^{-1}$. The components of the inverse metric are usually denoted as $g^{ij}$ where $g_{ij} g^{jk} = \delta^i_k$ is a defining property. The "index-raising" from forms to vectors is then defined by $$w^i = \beta_j g^{ij}$$ In fact, this operation is so common in metric geometry that physicists (especially relativists) simply say that it is "the same object with indices up or down", i.e. $$\kappa_i = v^j g_{ij} \equiv v_i$$ and $$w^i = \beta_j g^{ij} \equiv \beta^i$$


Non-orthogonal bases

We can take the contravariant basis vectors and check whether they are orthogonal by taking simply $$\mathbf{g}(\mathbf{e}_{(i)},\mathbf{e}_{(j)}) =?$$ or the covariant basis $$\mathbf{g}^{-1}(\mathbf{\epsilon}^{(i)},\mathbf{\epsilon}^{(j)}) =?$$ In this sense, covariant and contravariant basis vectors can be orthogonal or non-orthogonal.

I am slightly uncomfortable with the way the article above "mixes" the covariant and contravariant vectors together. Trying to assess whether $\mathbf{e}_{(i)}$ and $\epsilon^{(j)}$ are orthogonal does not make a lot of sense because they are simply different geometrical objects. (If you try to raise/lower an index on either of them via the metric and then using the metric to dot-product the result, you will get orthogonality trivially from definitory properties of the bases and the metric/inverse metric cited above. You can easily verify that yourself.)

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  • $\begingroup$ thanks for the detailed answer. But I am not sure whether it answers the question that I asked. I am a newbie to differential geometry. Can you clarify ? $\endgroup$
    – user66043
    Commented Jan 31, 2016 at 14:36
  • $\begingroup$ Can you please explain how we are getting $κ_i=g_{ij}u^j$ from the equation $κ(w)≡g(u,w)$? Also can you name the reference from where you studied this? $\endgroup$ Commented Sep 14, 2023 at 18:49

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