In the context of non-orthogonal bases, I am trying to understand the mathematical relation between the metric tensor and the transformation matrix between the orthogonal and non-orthogonal basis.
I have seen this equation here:
From this, the author infers that:
I am lost with the notation and the indexes. I think that I would understand it with a specific example. Say you have two bases, blue and red, like this:
I can build the metric tensor matrix as a “total war” between the basis vectors of the red (non-orthonormal) basis, but expressing its moduli as measured by the blue basis, as follows:
$${g_{ij}} = \left[ {\begin{array}{*{20}{c}}{{g_{11}}}&{{g_{12}}}\\{{g_{21}}}&{{g_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{{\left[ {{{\left| {{{\vec e}_1}} \right|}^2}} \right]}_{blue}}}&{{{\left[ {\left| {{{\vec e}_1}} \right|} \right]}_{blue}}{{\left[ {\left| {{{\vec e}_2}} \right|} \right]}_{blue}}\cos {{45}^o}}\\{{{\left[ {\left| {{{\vec e}_1}} \right|} \right]}_{blue}}{{\left[ {\left| {{{\vec e}_2}} \right|} \right]}_{blue}}\cos {{45}^o}}&{{{\left[ {{{\left| {{{\vec e}_2}} \right|}^2}} \right]}_{blue}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1*1}&{1\sqrt 2 \frac{1}{{\sqrt 2 }}}\\{1\sqrt 2 \frac{1}{{\sqrt 2 }}}&{\sqrt 2 \sqrt 2 }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\1&2\end{array}} \right]
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$$
In turn, the transformation matrices would be as follows:
From red into blue: blue measures that red basis vectors have coordinates (1,0) and (1,1), which (if we put these as column vectors of a matrix) renders:
$$\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]
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$$
We can check that this indeed works to transform the red coordinated of red basis vectors into blue values:
$$\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\left( \begin{array}{l}1\\0\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&0\end{array}} \right)\\\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\left( \begin{array}{l}0\\1\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&1\end{array}} \right)\end{array}
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$$
To transform from blue into red, we have to invert the matrix:
$${\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]^{ – 1}} = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]
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$$
We can check that this indeed works:
$$\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]\left( \begin{array}{l}1\\0\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&0\end{array}} \right)\\\left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]\left( \begin{array}{l}0\\1\end{array} \right) = \left( {\begin{array}{*{20}{c}}{ – 1}&1\end{array}} \right)\end{array}
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$$
Now we see that multiplying transformation matrix into blue (transposed) by itself (not transposed) works (it gives the metric tensor):
$${\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]^T}\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1*1 + 0*0}&{1*1 + 0*1}\\{1*1 + 1*0}&{1*1 + 1*1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\1&2\end{array}} \right]
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$$
However, the equation at the beginning of the post seems to be using the transformation-into-red matrix, since it applies it to the blue or orthogonal basis vectors, whose product is the Kronecker delta. But this does not seem to work in my example,
no matter if we multiply transformation matrix into red (transposed) by itself (not transposed):
$${\left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]^T}\left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ – 1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1*1 + 0*0}&{1*( – 1) + 0*1}\\{( – 1)*1 + 1*0}&{( – 1)*( – 1) + 1*1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\{ – 1}&2\end{array}} \right]
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$$
or if we take the inverse order, as proposed by the said equation:
$$\left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]{\left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}1&{ – 1}\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\{ – 1}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1*1 + ( – 1)*( – 1)}&{1*0 + ( – 1)*1}\\{0*1 + 1*( – 1)}&{0*0 + 1*1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ – 1}\\{ – 1}&1\end{array}} \right]
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$$
So my questions are:
a) Is the equation copied at the start right?
b) In any case, could you explain what the right equation does in long form based on my numerical example?
Edit: I corrected the matrix products, where there were mistakes; now some results look more similar to the metric tensor, but still not.
Edit2: Following the idea in Kurt's answer of focusing on one entry of the metric tensor, I have put it this way, which might be closer to my original idea:
Best Answer
When we decompose a metric $g$ as $$ \underbrace{ \begin{bmatrix} 1&0\\1&1 \end{bmatrix}}_{\textstyle C} \underbrace{\begin{bmatrix} 1&1\\0&1 \end{bmatrix}}_{\textstyle C^\top}=\underbrace{\begin{bmatrix} 1&1\\1&2 \end{bmatrix}}_{\textstyle g} $$ Then we can create from the orthonormal basis $\mathbf{e}_1=\begin{bmatrix}1\\0\end{bmatrix},\mathbf{e}_2=\begin{bmatrix}0\\1\end{bmatrix}$ a new basis \begin{align} \mathbf{g}_1&=C_{11}\mathbf{e}_1+C_{12}\mathbf{e}_2= \mathbf{e}_1=\begin{bmatrix}1\\0\end{bmatrix}\,,\\ \mathbf{g}_2&=C_{21}\mathbf{e}_1+C_{22}\mathbf{e}_2= \mathbf{e}_1+\mathbf{e}_2=\begin{bmatrix}1\\1\end{bmatrix}\, \end{align} that it is not orthonormal. Since the dot products of those new basis vectors are $$ \mathbf{g}_i\cdot \mathbf{g}_j=\begin{bmatrix} 1&1\\1&2 \end{bmatrix} $$ they recover the original metric. In other words:
The matrix between $\mathbf{g}_i$ and $\mathbf{g}_j$ that represents that metric in the new basis is just the identity matrix while between $\mathbf{e}_i$ and $\mathbf{e}_j$ it is the matrix $g\,.$ For example: \begin{align} \underbrace{\begin{bmatrix}1&1\end{bmatrix}}_{\textstyle\mathbf{g}_2^\top} \begin{bmatrix}1&0\\0&1\end{bmatrix}\underbrace{\begin{bmatrix}1\\1\end{bmatrix}}_{\textstyle\mathbf{g}_2}= \underbrace{\begin{bmatrix}0&1\end{bmatrix}}_{\textstyle\mathbf{e}_2^\top} \begin{bmatrix}1&1\\1&2\end{bmatrix}\underbrace{\begin{bmatrix}0\\1\end{bmatrix}}_{\textstyle\mathbf{e}_2}\,. \end{align} Both sides of this equal $2=g_{22}\,.$ That's all.