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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:

enter image description here

From this, the author infers that:

enter image description here

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:

enter image description here

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] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaG8bcaWGNb % WaaSbaaSqaaiaadMgacaWGQbaabeaakiabg2da9maadmaabaqbaeqa % biGaaaqaaiaadEgadaWgaaWcbaGaaGymaiaaigdaaeqaaaGcbaGaam % 4zamaaBaaaleaacaaIXaGaaGOmaaqabaaakeaacaWGNbWaaSbaaSqa % aiaaikdacaaIXaaabeaaaOqaaiaadEgadaWgaaWcbaGaaGOmaiaaik % daaeqaaaaaaOGaay5waiaaw2faaiabg2da9maadmaabaqbaeqabiGa % aaqaamaadmaabaWaaqWaaabbOpaaaaaasvgza8qabaGabmyzayaala % WaaSbaaSqaaiaaigdaaeqaaaGcpaGaay5bSlaawIa7amaaCaaaleqa % baGaaGOmaaaaaOGaay5waiaaw2faaabbaaaaaG+acXwDLbWdcmaaBa % aaleaacaWGIbGaamiBaiaadwhacaWGLbaabeaaaOWdaeaadaWadaqa % amaaemaapeqaaiqadwgagaWcamaaBaaaleaacaaIXaaabeaaaOWdai % aawEa7caGLiWoaaiaawUfacaGLDbaapiWaaSbaaSqaaiaadkgacaWG % SbGaamyDaiaadwgaaeqaaOWdamaadmaabaWaaqWaa8qabaGabmyzay % aalaWaaSbaaSqaaiaaikdaaeqaaaGcpaGaay5bSlaawIa7aaGaay5w % aiaaw2faa8GadaWgaaWcbaGaamOyaiaadYgacaWG1bGaamyzaaqaba % GcpaGaci4yaiaac+gacaGGZbGaaGinaiaaiwdadaahaaWcbeqaaiaa % d+gaaaaakeaadaWadaqaamaaemaapeqaaiqadwgagaWcamaaBaaale % aacaaIXaaabeaaaOWdaiaawEa7caGLiWoaaiaawUfacaGLDbaapiWa % aSbaaSqaaiaadkgacaWGSbGaamyDaiaadwgaaeqaaOWdamaadmaaba % WaaqWaa8qabaGabmyzayaalaWaaSbaaSqaaiaaikdaaeqaaaGcpaGa % ay5bSlaawIa7aaGaay5waiaaw2faa8GadaWgaaWcbaGaamOyaiaadY % gacaWG1bGaamyzaaqabaGcpaGaci4yaiaac+gacaGGZbGaaGinaiaa % iwdadaahaaWcbeqaaiaad+gaaaaakeaadaWadaqaamaaemaapeqaai % qadwgagaWcamaaBaaaleaacaaIYaaabeaaaOWdaiaawEa7caGLiWoa % daahaaWcbeqaaiaaikdaaaaakiaawUfacaGLDbaapiWaaSbaaSqaai % aadkgacaWGSbGaamyDaiaadwgaaeqaaaaaaOWdaiaawUfacaGLDbaa % cqGH9aqpdaWadaqaauaabeqaciaaaeaacaaIXaGaaiOkaiaaigdaae % aacaaIXaWaaOaaaeaacaaIYaaaleqaaOWaaSaaaeaacaaIXaaabaWa % aOaaaeaacaaIYaaaleqaaaaaaOqaaiaaigdadaGcaaqaaiaaikdaaS % qabaGcdaWcaaqaaiaaigdaaeaadaGcaaqaaiaaikdaaSqabaaaaaGc % baWaaOaaaeaacaaIYaaaleqaaOWaaOaaaeaacaaIYaaaleqaaaaaaO % Gaay5waiaaw2faaiabg2da9maadmaabaqbaeqabiGaaaqaaiaaigda % aeaacaaIXaaabaGaaGymaaqaaiaaikdaaaaacaGLBbGaayzxaaaaaa!B25A! $$

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:

$$\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaG8bdaWada % aeeaaaaaa6dieB1vgapeqaauaabeqaciaaaeaacaaIXaaabaGaaGym % aaqaaiaaicdaaeaacaaIXaaaaaWdaiaawUfacaGLDbaaaaa!3E5B! $$

We can check that this indeed works to transform the red coordinated of red basis vectors into blue values:

$$\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\left( \begin{array}{l}1\\0\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&0\end{array}} \right)\\\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]\left( \begin{array}{l}0\\1\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&1\end{array}} \right)\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceiqabeaaiFqaam % aadmaaqqaaaaaaOpGqSvxza8qabaqbaeqabiGaaaqaaiaaigdaaeaa % caaIXaaabaGaaGimaaqaaiaaigdaaaaapaGaay5waiaaw2faamaabm % aaeaqababbOpaaaaaasvgza8GabaGaaGymaaqaaiaaicdaaaWdaiaa % wIcacaGLPaaacqGH9aqpqaaaaaaaaaWddmaabmaapeqaauaabeqabi % aaaeaacaaIXaaabaGaaGimaaaaa8WacaGLOaGaayzkaaaapaqaamaa % dmaapeqaauaabeqaciaaaeaacaaIXaaabaGaaGymaaqaaiaaicdaae % aacaaIXaaaaaWdaiaawUfacaGLDbaadaqadaabaeqapiqaaiaaicda % aeaacaaIXaaaa8aacaGLOaGaayzkaaGaeyypa0Zddmaabmaapeqaau % aabeqabiaaaeaacaaIXaaabaGaaGymaaaaa8WacaGLOaGaayzkaaaa % aaa!547C! $$

To transform from blue into red, we have to invert the matrix:

$${\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaG8bdaWada % aeeaaaaaa6dieB1vgapeqaauaabeqaciaaaeaacaaIXaaabaGaaGym % aaqaaiaaicdaaeaacaaIXaaaaaWdaiaawUfacaGLDbaadaahaaWcbe % qaaiabgkHiTiaaigdaaaGccqGH9aqpdaWadaaeeG+aaaaaaivzKbWd % ceaafaqabeGacaaabaGaaGymaaqaaiabgkHiTiaaigdaaeaacaaIWa % aabaGaaGymaaaaa8aacaGLBbGaayzxaaaaaa!4954! $$

We can check that this indeed works:

$$\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right]\left( \begin{array}{l}1\\0\end{array} \right) = \left( {\begin{array}{*{20}{c}}1&0\end{array}} \right)\\\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right]\left( \begin{array}{l}0\\1\end{array} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}&1\end{array}} \right)\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceiqabeaaiFqaam % aadmaaqqa6daaaaaGuLrgapeqaauaabeqaciaaaeaacaaIXaaabaGa % eyOeI0IaaGymaaqaaiaaicdaaeaacaaIXaaaaaWdaiaawUfacaGLDb % aadaqadaabaeqapeqaaabbaaaaaG+acXwDLbWdciaaigdaa8qabaWd % ciaaicdaaaWdaiaawIcacaGLPaaacqGH9aqpqaaaaaaaaaWddmaabm % aapeqaauaabeqabiaaaeaacaaIXaaabaGaaGimaaaaa8WacaGLOaGa % ayzkaaaapaqaamaadmaapeqaauaabeqaciaaaeaacaaIXaaabaGaey % OeI0IaaGymaaqaaiaaicdaaeaacaaIXaaaaaWdaiaawUfacaGLDbaa % daqadaabaeqapeqaa8GacaaIWaaabaGaaGymaaaapaGaayjkaiaawM % caaiabg2da98WadaqadaWdbeaafaqabeqacaaabaGaeyOeI0IaaGym % aaqaaiaaigdaaaaapmGaayjkaiaawMcaaaaaaa!5784! $$

Now we see that multiplying transformation matrix into blue (transposed) by itself (not transposed) works (it gives the metric tensor):

$${\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right]^T}\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1*1 + 0*0}&{1*1 + 0*1}\\{1*1 + 1*0}&{1*1 + 1*1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&1\\1&2\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaG8bdaWada % aeeaaaaaa6dieB1vgapeqaauaabeqaciaaaeaacaaIXaaabaGaaGym % aaqaaiaaicdaaeaacaaIXaaaaaWdaiaawUfacaGLDbaadaahaaWcbe % qaaiaadsfaaaGcdaWadaWdbeaafaqabeGacaaabaGaaGymaaqaaiaa % igdaaeaacaaIWaaabaGaaGymaaaaa8aacaGLBbGaayzxaaGaeyypa0 % ZaamWaa8qabaqbaeqabiGaaaqaaiaaigdaaeaacaaIWaaabaGaaGym % aaqaaiaaigdaaaaapaGaay5waiaaw2faamaadmaapeqaauaabeqaci % aaaeaacaaIXaaabaGaaGymaaqaaiaaicdaaeaacaaIXaaaaaWdaiaa % wUfacaGLDbaacqGH9aqpdaWadaWdbeaafaqabeGacaaabaGaaGymai % aacQcacaaIXaGaey4kaSIaaGimaiaacQcacaaIWaaabaGaaGymaiaa % cQcacaaIXaGaey4kaSIaaGimaiaacQcacaaIXaaabaGaaGymaiaacQ % cacaaIXaGaey4kaSIaaGymaiaacQcacaaIWaaabaGaaGymaiaacQca % caaIXaGaey4kaSIaaGymaiaacQcacaaIXaaaaaWdaiaawUfacaGLDb % aacqGH9aqpdaWadaqaauaabeqaciaaaeaacaaIXaaabaGaaGymaaqa % aiaaigdaaeaacaaIYaaaaaGaay5waiaaw2faaaaa!6D55! $$

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):

$${\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right]^T}\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\{ - 1}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1*1 + 0*0}&{1*( - 1) + 0*1}\\{( - 1)*1 + 1*0}&{( - 1)*( - 1) + 1*1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaG8bdaWada % aeeG+aaaaaaivzKbWdbeaafaqabeGacaaabaGaaGymaaqaaiabgkHi % TiaaigdaaeaacaaIWaaabaGaaGymaaaaa8aacaGLBbGaayzxaaWaaW % baaSqabeaacaWGubaaaOWaamWaa8qabaqbaeqabiGaaaqaaiaaigda % aeaacqGHsislcaaIXaaabaGaaGimaaqaaiaaigdaaaaapaGaay5wai % aaw2faaiabg2da9maadmaapeqaauaabeqaciaaaeaacaaIXaaabaGa % aGimaaqaaiabgkHiTiaaigdaaeaacaaIXaaaaaWdaiaawUfacaGLDb % aadaWadaWdbeaafaqabeGacaaabaGaaGymaaqaaiabgkHiTiaaigda % aeaacaaIWaaabaGaaGymaaaaa8aacaGLBbGaayzxaaGaeyypa0Zaam % Waa8qabaqbaeqabiGaaaqaaiaaigdacaGGQaGaaGymaiabgUcaRiaa % icdacaGGQaGaaGimaaqaaiaaigdacaGGQaGaaiikaiabgkHiTiaaig % dacaGGPaGaey4kaSIaaGimaiaacQcacaaIXaaabaGaaiikaiabgkHi % TiaaigdacaGGPaGaaiOkaiaaigdacqGHRaWkcaaIXaGaaiOkaiaaic % daaeaacaGGOaGaeyOeI0IaaGymaiaacMcacaGGQaGaaiikaiabgkHi % TiaaigdacaGGPaGaey4kaSIaaGymaiaacQcacaaIXaaaaaWdaiaawU % facaGLDbaacqGH9aqpdaWadaqaauaabeqaciaaaeaacaaIXaaabaGa % eyOeI0IaaGymaaqaaiabgkHiTiaaigdaaeaacaaIYaaaaaGaay5wai % aaw2faaaaa!7B8E! $$

or if we take the inverse order, as proposed by the said equation:

$$\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right]{\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0\\{ - 1}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1*1 + ( - 1)*( - 1)}&{1*0 + ( - 1)*1}\\{0*1 + 1*( - 1)}&{0*0 + 1*1}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 1}&1\end{array}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaG8bdaWada % aeeG+aaaaaaivzKbWdbeaafaqabeGacaaabaGaaGymaaqaaiabgkHi % TiaaigdaaeaacaaIWaaabaGaaGymaaaaa8aacaGLBbGaayzxaaWaam % Waa8qabaqbaeqabiGaaaqaaiaaigdaaeaacqGHsislcaaIXaaabaGa % aGimaaqaaiaaigdaaaaapaGaay5waiaaw2faamaaCaaaleqabaGaam % ivaaaakiabg2da9maadmaapeqaauaabeqaciaaaeaacaaIXaaabaGa % eyOeI0IaaGymaaqaaiaaicdaaeaacaaIXaaaaaWdaiaawUfacaGLDb % aadaWadaWdbeaafaqabeGacaaabaGaaGymaaqaaiaaicdaaeaacqGH % sislcaaIXaaabaGaaGymaaaaa8aacaGLBbGaayzxaaGaeyypa0Zaam % Waa8qabaqbaeqabiGaaaqaaiaaigdacaGGQaGaaGymaiabgUcaRiaa % cIcacqGHsislcaaIXaGaaiykaiaacQcacaGGOaGaeyOeI0IaaGymai % aacMcaaeaacaaIXaGaaiOkaiaaicdacqGHRaWkcaGGOaGaeyOeI0Ia % aGymaiaacMcacaGGQaGaaGymaaqaaiaaicdacaGGQaGaaGymaiabgU % caRiaaigdacaGGQaGaaiikaiabgkHiTiaaigdacaGGPaaabaGaaGim % aiaacQcacaaIWaGaey4kaSIaaGymaiaacQcacaaIXaaaaaWdaiaawU % facaGLDbaacqGH9aqpdaWadaqaauaabeqaciaaaeaacaaIYaaabaGa % eyOeI0IaaGymaaqaaiabgkHiTiaaigdaaeaacaaIXaaaaaGaay5wai % aaw2faaaaa!7B8E! $$

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:

enter image description here

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7
  • $\begingroup$ The equation you have seen $g_{ij}=...=C_{i\ell}C_{j\ell}$ seems clear to you (is it?). It is the matrix multiplication $[g]=[C][C]^\top$ in index notation. I find the author's way to write this as $[g_{ij}]=[C_{ij}][C_{ij}]^\top$ to be sloppy, if not misleading to the point of being wrong. Again: $g$ is the matrix product, not a Hadamard product. $\endgroup$
    – Kurt G.
    Jul 28, 2023 at 7:45
  • $\begingroup$ @Kurt G. It is "the matrix product", but which matrix? Transformation matrix into blue or orthonormal (which in my example works, but with the oppposite order, i.e. transposed first) or into read or oblique (which is what it seems, but in my example, if I did it well, would not work, no matter the order)? $\endgroup$
    – Sierra
    Jul 28, 2023 at 9:36
  • $\begingroup$ Is it not obvious from $$ \underbrace{ \begin{bmatrix} 1&1\\0&1 \end{bmatrix}^\top}_{\textstyle C} \underbrace{\begin{bmatrix} 1&1\\0&1 \end{bmatrix}}_{\textstyle C^\top}=\underbrace{\begin{bmatrix} 1&1\\1&2 \end{bmatrix}}_{\textstyle g} $$ what the matrices are? $\endgroup$
    – Kurt G.
    Jul 28, 2023 at 15:48
  • $\begingroup$ It is obvious to me that this expression is the transformation-into-blue (into orthonormal basis) matrix, left-multiplied by its transpose. Let us call this M^T*M. If the thing is that the author is defining M^T as C, that would look to me confusing (why so?) but no problem. But then still the question is why these matrices (transforming into blue = orthonormal) are applied to the e basis vector = the blue ones = the orthonormal ones... $\endgroup$
    – Sierra
    Jul 28, 2023 at 17:48
  • $\begingroup$ Sounds like we agreed that the metric $g$ can be decomposed as $g=CC^\top\,.$ The author transforms the orthonormal basis vectors $\mathbf{e}_i$ via $\mathbf{g}_j=C_{j\ell}\mathbf{e}_\ell$ into a new basis. In that (non orthonormal) basis the same metric is obviously represented by the identity matrix: $g_{ij}=\mathbf{g}_i\cdot \mathbf{g}_j\,.$ Confusing overuse of $g$ for metric and basis but that's all. $\endgroup$
    – Kurt G.
    Jul 28, 2023 at 18:18

1 Answer 1

1
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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.

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5
  • $\begingroup$ It is a very helpful idea to focuse on a particular entry of the metric tensor. Since I understand it better based on what I called M (instead of C) in my comments, I have attempted at developing a version of the same entry as an Edit of the question. Maybe I would just need a little push to generalize that into a formula embracing the whole metric tensor. $\endgroup$
    – Sierra
    Jul 30, 2023 at 5:43
  • $\begingroup$ Clearly there is an $M^T$ and an $M$. The blue part is the Kronecker delta. Can we say that the red part (coordinates of the oblique basis vectors in their own language) is also Kronecker? $\endgroup$
    – Sierra
    Jul 30, 2023 at 5:52
  • $\begingroup$ "Red, blue, $C,C^\top,M,M^\top$" ... and what not. You are overthinking this or get lost in your own overcomplex notation. Please focus on my answer and ask specific questions about that. $\endgroup$
    – Kurt G.
    Jul 30, 2023 at 8:38
  • $\begingroup$ You are right. Focusing then on the left-hand side of your last equation, it seems that $g_2^T$ generalizes into $C$ and $g_2$ into $C^T$, and that is it, right? $\endgroup$
    – Sierra
    Jul 30, 2023 at 12:59
  • $\begingroup$ I don't know that it means that a vector generalizes into a matrix. If it helps here are the relations again: $\mathbf{g}_i=C_{i\ell}\,\mathbf{e}_\ell\,,\;\mathbf{g}_i^\top=\mathbf{e}_\ell^\top C_{\ell i}^\top\,.$ $\endgroup$
    – Kurt G.
    Jul 30, 2023 at 15:38

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