Thank you for letting me know that.

Thank you for letting me know that. It is much appreciated! Cheers!

Thank you so much for those kind words. It is much appreciated. I will look at the Petrov classifications but it may be some time before I do so. Best wishes with your studies. Cheers!

@@TensorCalculusRobertDavie Thank you so much ❤️

You're welcome.

Thank you for that feedback. Cheers!

Thank you for that!

You're welcome.

Hello and thank you for your comment.

Thank you for saying that.

Thank you!

Cheers!

Thank you for saying that.

You're welcome!

Which means it is not 0.

Thank you for pointing that out. I have added the following to the video.🙂🙂🙂 Correction: The second last line in the argument at 8:05 should read, 1/2(ω_12 dx^1∧dx^2 + ω_12 dx^1∧dx^2). Since, ω_12 = − ω_21 and dx^1∧dx^2 = − dx^2∧dx^1 combine to produce a positive sign in the expression for this line.

Thank you for your enthusiastic comment! It's wonderful to hear your appreciation for 𝑘-forms-these objects are indeed incredibly powerful tools in differential geometry and mathematical physics. Key Insight on 𝑘-Forms While 𝑘-forms span an essential subspace of the tensor space, particularly the alternating multilinear forms, it’s worth noting a few clarifying points: 1. Multi-tensor Space: The full multi-tensor space includes both symmetric and antisymmetric components, while 𝑘-forms specifically span the antisymmetric part. The basis for 𝑘-forms consists of wedge products of differentials like 𝑑𝑥^𝑖 ∧ 𝑑𝑥^𝑗 , capturing oriented volume elements at each point of a manifold. 2. Algebraic Structure: The exterior algebra built from 𝑘-forms gives a natural framework for many geometric constructions. Even though 𝑘-forms do not directly span the entire tensor space, they are fundamental for expressing ideas such as flux, circulation, and integration over submanifolds. 3. Decomposition: Multi-tensors can often be decomposed into symmetric and antisymmetric components, where the antisymmetric portion relates directly to 𝑘-forms. This connection reinforces how central 𝑘-forms are in structuring more general tensor spaces. Your excitement is well-placed-𝑘-forms provide a compact, elegant way to express complex geometric and physical phenomena. Thanks for your thoughtful engagement with the material!

Thank you for your detailed and insightful questions! Let’s address them step by step: 1. Why are partial derivatives equal to basis vectors? This is a foundational concept in differential geometry, so let’s break it down. Partial derivatives ∂/∂𝑥^𝑖 are interpreted as basis vectors in the tangent space because of how they act on scalar functions. Specifically: - The tangent space at a point 𝑝 on a manifold is the space of all possible "directions" in which one can move away from 𝑝. - A vector in this tangent space is defined operationally as a differential operator that acts on smooth functions 𝑓 defined near 𝑝. In local coordinates (𝑥^1, 𝑥^2 , … , 𝑥^𝑛), the operators ∂/∂𝑥^𝑖 naturally arise from infinitesimal variations in each coordinate. To "prove" this equivalence, consider: - The definition of ∂/∂𝑥^𝑖 as a directional derivative: it measures the rate of change of a function 𝑓 along the 𝑥^𝑖-axis. - In the coordinate basis {∂/∂𝑥^1 , ∂/∂𝑥^2 , … , ∂/∂𝑥^𝑛} , any tangent vector 𝑣 can be written as a linear combination 𝑣 = 𝑣^𝑖 ∂/∂𝑥^𝑖 . The coefficients 𝑣^𝑖 correspond to components of the vector, and the basis vectors are precisely the partial derivative operators. Where your example went astray: The mistake lies in interpreting the action of 𝑒_𝑥 (a basis vector) as a dot product. Instead: - 𝑒_𝑥 , as ∂/∂𝑥 , acts on 𝑓(𝑥,𝑦) by taking its derivative with respect to 𝑥, yielding 2𝑥 for your example 𝑓(𝑥,𝑦) = 𝑥^2 + 𝑦^2 . - The dot product interpretation doesn’t apply here because 𝑒_𝑥 (as ∂/∂𝑥) is not a vector in the traditional Euclidean sense; it’s a differential operator. 2. Is there a simple general proof? Here’s a conceptual outline: i. Define a vector in the tangent space 𝑇_𝑝(𝑀) at 𝑝 as a derivation: a linear map 𝑣 : 𝐶^∞(𝑀) → 𝑅 satisfying the Leibniz rule 𝑣(𝑓𝑔) = 𝑣(𝑓)𝑔 + 𝑓𝑣(𝑔). ii. In local coordinates, show that the partial derivatives ∂/∂𝑥^𝑖 satisfy these properties and form a basis for all derivations at 𝑝. iii. Conclude that the partial derivatives ∂/∂𝑥^𝑖 serve as the coordinate basis for the tangent space. 3. About the curve at 1:30 You’re asking whether the curve is defined relative to an origin off the surface or on it. Great question! - Generally, a curve on a manifold is defined intrinsically, meaning it doesn’t depend on an "external" origin. It’s described as a mapping 𝛾 : 𝐼 → 𝑀 , where 𝐼 ⊂ 𝑅 is an interval and 𝑀 is the manifold. - If you’re working in an embedded setting (e.g., the surface exists in 𝑅^3), you can think of the curve as being described by a position vector relative to some origin in 𝑅^3. However, the intrinsic description relies only on the coordinates within the surface itself, so the "origin" would be irrelevant or purely local. Summary 1. Partial derivatives are basis vectors in the tangent space because they naturally represent infinitesimal motions along coordinate directions. 2. Your example misinterpreted ∂/∂𝑥 as a dot product; it’s instead an operator that acts on functions. 3. The curve can be understood intrinsically without needing an origin outside the surface. Thanks again for your questions, and I hope this clarifies things! Feel free to reach out if you’d like further discussion.

I appreaciate your response I feel like i sort of am beginning to gain an intuition about it, the partial derivative operator when acting on a function tells the change in the function value with respect to a change / step in a particular direction (𝑥^𝑖), and what else takes a step in a particular direction? well a basis vector.. I am following all up until 7:40, where we get d/dλ = t^μ ∂/∂x^μ, (i agree with this), but then you're mentioning the earlier definition of a tangent vector t = t^μ e_μ, and since the components match and the derivative operators measures in same direction that d/dλ = t, but can we really say that when we have 1 variable being the same and having 3 total variables in the equation?, to illustrate what i mean here: 6 = 3 * 2 and 9 = 3 * 3 where 3 is the common factor like t^μ. Now i'm not a pure math or physics student, so i don't quite understand the formal proof. I'm also quite new to (differential geometry, so i don't have the best knowledge about all the concepts).

Thank you for your thoughtful question! Let’s break this down step by step to clarify why we use the determinant of the Jacobian rather than the determinant of the inverse Jacobian when working with pullbacks. Area Scaling and the Pullback When a function 𝑓:𝑀→𝑁 maps a region in 𝑀 to a region in 𝑁, the Jacobian determinant of 𝑓 at a point tells us the local factor by which areas (or volumes in higher dimensions) are scaled under the mapping 𝑓. For instance: If det(𝐽_𝑓) = 6 , this means that a small area in 𝑀 is stretched to an area 6 times larger in 𝑁. The pullback process operates in a different way. Instead of physically "mapping areas back" from 𝑁 to 𝑀, the pullback transforms differential forms. This transformation ensures that integrals of forms over corresponding regions remain consistent. Why the Jacobian Determinant is Used To see why the Jacobian determinant (not its inverse) is used for pullbacks, recall how integrals transform under a change of variables: ∫_𝑁 𝜔 = ∫_𝑀 𝑓^∗(𝜔) , where 𝜔 is a differential form on 𝑁, and 𝑓^∗(𝜔) is its pullback to 𝑀. 1. Change of Variables Formula: When we change variables in an integral, the transformation involves multiplying by ∣det(𝐽_𝑓)∣. This factor accounts for how 𝑓 scales the "infinitesimal volume elements" during the mapping from 𝑀 to 𝑁. 2. Pullback Transformation: The pullback 𝑓^∗(𝜔) must ensure that the integral over 𝑀 gives the same result as the integral over 𝑁. For this to work, the pullback introduces the same scaling factor ∣det(𝐽_𝑓)∣ , not 1 / ∣det(𝐽_𝑓)∣. The reason is that while the pullback “pulls the form back,” the integral over 𝑀 still requires us to consider how the geometry of 𝑀 maps under 𝑓. Why Not the Determinant of the Inverse Jacobian? The inverse Jacobian is relevant if you are directly "reversing" the transformation, as in finding 𝑓^−1. However, the pullback 𝑓^∗ doesn't work this way-it doesn't compute the inverse transformation but instead ensures that integrals remain consistent. The determinant of the Jacobian reflects the local scaling of areas (or volumes) induced by 𝑓, which is precisely what the pullback must account for. Intuitive Answer to Your Example If 𝑓 scales areas by a factor of 6, you might think the pullback should "scale by 1/6" because it's reversing the direction. But instead, the pullback ensures that the "area differential" (or volume form) in 𝑀 transforms correctly so the integral gives the same result as it would on 𝑁. This is why we still use the determinant of the Jacobian, not its inverse. I hope this explanation clarifies things! Let me know if you'd like to dig deeper into the topic or discuss specific examples. Thanks again for engaging with the video!

@@TensorCalculusRobertDavie Okay, I understand that the pullback ensures that the integrals remain consistent, but why is it the Jacobian and not its inverse that keeps it consistent? Is it because the "differentials" that 𝑓* acts on transform contra-variantly from the vectors that 𝑓 acts on? So it's like a double-inverse situation?

@@APaleDot It’s a great insight to connect the behavior of the pullback with the contravariant transformation of differentials. Let me clarify why it is the Jacobian and not its inverse that keeps the integrals consistent. Key Ideas: 1. The Pullback and Its Action on Differential Forms: The pullback 𝑓^∗ maps differential forms on the target manifold 𝑁 back to the source manifold 𝑀. Specifically, if 𝜔 is a differential form on 𝑁, the pullback 𝑓^∗(𝜔) is a form on 𝑀. - Differential forms, like 𝑑𝑥∧𝑑𝑦, transform contravariantly compared to vectors. This means their transformation involves the Jacobian matrix directly, not its inverse. 2. Why the Jacobian Appears: - When integrating a differential form over a manifold, the integral depends on how the coordinate system transforms. For a map 𝑓 : 𝑀 → 𝑁, the pullback 𝑓^∗(𝜔) effectively "adjusts" the form so that the integral of 𝜔 over 𝑓(𝑀) (a region in 𝑁) matches the integral of 𝑓^∗(𝜔) over 𝑀. - The Jacobian determinant arises because it scales areas, volumes, or higher-dimensional measures when transitioning between coordinate systems. The pullback ensures the integral remains consistent under this scaling. 3. Why Not the Inverse Jacobian?: - The Jacobian determinant scales the "volume" element 𝑑𝑥^1 ∧ 𝑑𝑥^2 ∧ … of the source manifold 𝑀 to align it with the volume element on 𝑁 as mapped by 𝑓. - The inverse Jacobian is associated with the forward transformation of vectors (contravariant objects), not the pullback of forms (covariant objects). In essence, the pullback already accounts for the contravariant transformation, so the Jacobian determinant is the correct scaling factor. 4. "Double-Inverse" Interpretation: Your intuition about a "double-inverse situation" is on the right track. Here’s why: - The pullback 𝑓^∗ operates on differential forms, which transform contravariantly. - The Jacobian determinant arises naturally in the transformation rule for forms, ensuring consistency with the way areas, volumes, and other measures change under 𝑓. - If you were instead transforming vectors or tangent vectors (contravariant objects), the inverse Jacobian would appear. Summary: The reason the Jacobian determinant appears, rather than the inverse Jacobian, is tied to the nature of differential forms and how they transform under pullbacks. Forms are covariant objects, so their transformation under 𝑓^∗ involves the direct Jacobian, not its inverse. Your mention of the "contra-variant transformation of differentials" is absolutely correct and aligns with this reasoning! I hope this clears things up. Let me know if you’d like further clarification or examples to explore this idea more deeply!

@@TensorCalculusRobertDavie Is this AI generated?

You're welcome.

Thank you!

Thank you for your thoughtful comment and keen observations! You’re absolutely correct that differential forms have tensorial properties, and this fact underpins both the pullback operation and coordinate transformations in integration. Pullbacks and Coordinate Changes In your example, transitioning from Cartesian to polar coordinates for integration is conceptually equivalent to using a pullback from the manifold described in Cartesian coordinates to a representation in polar coordinates. The pullback essentially ensures that the integral of a differential form remains invariant by adjusting the form to account for the coordinate transformation. This invariance aligns with your intuition that differential forms transform as covariant tensors. Tensorial Transformation of Differential Forms Differential forms indeed behave as covariant tensors, specifically antisymmetric ones. For a differential 1-form 𝜔 = 𝑓(𝑥) 𝑑𝑥 in Cartesian coordinates, a change to polar coordinates involves applying the chain rule: 𝑑𝑥 = ∂𝑥/∂𝑟 𝑑𝑟 + ∂𝑥/∂𝜃 𝑑𝜃. This transformation embodies the same principles captured by the pullback operation, as the pullback formalism inherently accounts for these transformations when mapping between coordinate representations. Consistency of Results Yes, the same results are obtained whether you approach the problem using: 1. Pullback Formalism: A powerful geometric tool applicable to any smooth map between manifolds. 2. Tensor Transformation Rules: The classical method of transforming components under coordinate changes. The consistency stems from both approaches being fundamentally expressions of the same underlying geometric structure. Demonstrating the Connection Consider integrating a differential form over a region 𝑀. If you compute the integral in new coordinates via tensor transformation properties, the process boils down to expressing the form in the new basis (i.e., transforming components) and integrating accordingly. The pullback approach accomplishes this systematically by mapping the form to the transformed manifold representation, yielding equivalent results. Your question beautifully highlights how the abstract formalism of pullbacks ties to familiar tensor properties. I appreciate your curiosity and will hopefully explore ways to present this connection visually in future content. Thank you for engaging so thoughtfully.

Thank you for your thoughtful question! In the last line I left the summation symbol out in order to save space and keep everything on the one line. This seemed a reasonable choice given that ω was defined above. Also, I left the wedge product out for the same reason when it should really be present when describing an integral over a manifold.

Thank you for your question and for engaging with the video! The example at 16:05 does indeed show a non-zero Lie bracket of the two vector fields. However, this does not necessarily imply torsion on the manifold. The Lie bracket measures the failure of two vector fields to commute, and it is purely a property of the vector fields themselves and their relationship to the manifold's geometry. Specifically, it is a way of capturing how the flows generated by the vector fields interact. A non-zero Lie bracket simply means that the flows of the two vector fields "twist" around each other in some way. Torsion, on the other hand, is a concept that arises in the context of an affine connection, which is an additional structure you can define on a manifold. Torsion specifically measures the antisymmetric part of the connection and tells you how much the connection fails to be symmetric. In particular, the torsion tensor 𝑇(𝑋,𝑌) is defined as: 𝑇(𝑋,𝑌) = ∇_𝑋 𝑌 − ∇_𝑌 𝑋 − [𝑋,𝑌] , where ∇ is the connection. In the absence of an affine connection being defined, you cannot speak of torsion. A manifold can have non-zero Lie brackets for its vector fields and still have torsion-free connections (e.g., the Levi-Civita connection in Riemannian geometry is torsion-free by definition). In short: - A non-zero Lie bracket tells us about the structure of the vector fields and their flows. - Torsion depends on the connection and involves both the Lie bracket and how the connection behaves. So, the non-zero Lie bracket in the example does not necessarily mean the manifold has torsion. It is simply a property of the vector fields being considered. Thank you for the thoughtful question! Let me know if you’d like further clarification.

@@TensorCalculusRobertDavie thank you for your amazing explanation!

You raise an excellent point, and I appreciate the question! Let’s explore whether the limit in the formula at 12:00 resolves the issue. The Role of the Limit: The formula presented is: 𝐿_𝑋 𝑌 = lim_(𝑡→0) [𝜙_𝑡^∗(𝑌) − 𝑌] / 𝑡 , where: - 𝜙_𝑡^∗(𝑌) is the pullback of 𝑌 along the flow generated by 𝑋, - 𝑌 is the vector field being compared at 𝜙_0(𝑝), the point on the flow corresponding to 𝑡 = 0. When 𝑡 → 0: 1. 𝜙_𝑡(𝑝) → 𝜙_0(𝑝), and 2. The pullback 𝜙_𝑡^∗(𝑌) transforms 𝑌(𝜙_𝑡(𝑝)) into a vector in the tangent space at 𝜙_0(𝑝). Thus, even though 𝜙_𝑡(𝑝) ≠ 𝜙_0(𝑝) for 𝑡 ≠ 0, the pullback ensures that 𝜙_𝑡^∗(𝑌) is always "reinterpreted" in the tangent space at 𝜙_0(𝑝). Does the Limit Handle the Discrepancy? Yes, the limit inherently resolves the issue because: - The pullback 𝜙_𝑡^∗(𝑌) maps 𝑌(𝜙_𝑡(𝑝)) into the tangent space at 𝜙_0(𝑝) , ensuring that the subtraction 𝜙_𝑡^∗(𝑌) - 𝑌 is always valid. - As 𝑡 → 0 , 𝜙_𝑡^∗(𝑌) - 𝑌 , smoothly converging to the value of 𝑌 at 𝜙_0(𝑝). This means that the subtraction and the definition of the Lie derivative are well-defined for all 𝑡, not just in the limit. The process of pulling back guarantees that vectors are always compared in the same space. Why Mentioning 𝜙_0(𝑝) is Still Helpful: Your concern stems from the explicit appearance of 𝜙_𝑡(𝑝) in the second term of the numerator, which might suggest (incorrectly) that 𝑌(𝜙_𝑡(𝑝)) is being directly compared to 𝑌(𝜙_0(𝑝)) without accounting for the pullback. While the formula is correct as written, emphasizing that 𝜙_𝑡^∗(𝑌) always operates in the tangent space of 𝜙_0(𝑝) can help avoid confusion. The pullback mechanism "bridges the gap" between 𝜙_𝑡(𝑝) and 𝜙_0(𝑝) , but this subtlety might not be immediately apparent without a detailed explanation. Conclusion: The limit indeed resolves the issue, and the formula as presented is rigorously correct. However, explicitly stating how the pullback aligns vectors into the same tangent space can make the explanation clearer and address potential misunderstandings. Thank you for raising this nuanced question-it highlights an important aspect of the Lie derivative that deserves careful attention!

@@TensorCalculusRobertDavie Thank you, Robert, for taking the time and Happy New Year. I apologize that I need to ask further because while the concept is clear that the vectors need to be in the same space to be subtracted, the notation seems to indicate otherwise. (Personal note: Time off is over, I'm back to my day job and maybe not able to focus as clearly on math. My apologies in advance if I'm just confused.) Let me pinpoint the issue. In your response above you write: 𝐿_𝑋 𝑌 = lim_(𝑡→0) [𝜙_𝑡^∗(𝑌) − 𝑌] / 𝑡 as the formula for the Lie derivative. Yes, but this expression doesn't fully specify the points at which these Y vectors are defined. In the formula on the slide at 12:00, both are written as 𝑌(𝜙_𝑡(𝑝)), This means that the first Y, after pullback is in a different space than the second Y, which hasn't been pulled back and is in its original tangent space at 𝜙_𝑡(𝑝). It seems the second Y should be 𝑌(𝜙_0(𝑝)), not 𝑌(𝜙_t(𝑝)). Otherwise the difference quotient isn't defined for any given t so the limit operation is not valid. I think this may just be an issue with a subscript on one term in the formula. Maybe? Thanks again for your help.

Thank you for your question-it’s a very insightful one! At 12:31, the ordering of the coordinates in the wedge product (and therefore the integral) is indeed critical because differential forms, and their integrals, respect the rules of antisymmetry. Swapping 𝑥 and 𝑦 in the wedge product 𝑑𝑥 ∧ 𝑑𝑦 gives: 𝑑𝑥 ∧ 𝑑𝑦 = −𝑑𝑦 ∧ 𝑑𝑥. So, as you correctly noted, if you swapped 𝑥 and 𝑦, the integral would acquire a minus sign. To address the second part of your question: how do we know the correct order? The order of the wedge product (e.g., 𝑑𝑥 ∧ 𝑑𝑦) is determined by the orientation of the manifold or domain on which we are integrating. Here’s how it works: 1. Orientation of the domain: When we integrate a 2-form (like 𝑓(𝑥,𝑦)𝑑𝑥 ∧ 𝑑𝑦) over a 2-dimensional domain, the order 𝑑𝑥 ∧ 𝑑𝑦 reflects a chosen orientation for that domain. For example, in the standard orientation of the Cartesian plane, 𝑑𝑥 ∧ 𝑑𝑦 is typically considered positive. If you reversed the order to 𝑑𝑦 ∧ 𝑑𝑥, you would effectively be using the opposite orientation, which flips the sign of the integral. 2. Context in physics or geometry: In most cases, the orientation is either explicitly stated or implicitly understood based on conventions. For instance, in the plane, the "natural" orientation is usually 𝑑𝑥 ∧ 𝑑𝑦. In higher dimensions, orientation can depend on the coordinate chart or the problem setup, but the wedge product always ensures consistency. To summarize: - Swapping 𝑥 and 𝑦 in the wedge product 𝑑𝑥 ∧ 𝑑𝑦 changes the sign of the integral. - The correct order is determined by the orientation of the manifold or domain. If the orientation is reversed, the sign of the integral also reverses, but as long as we stick to the chosen orientation throughout the problem, everything remains consistent. Let me know if you’d like further clarification, and thanks again for your excellent question!

Thank you so much for your detailed comment and for engaging so thoughtfully with the material! I really appreciate your observations, as well as your suggestions for making the explanation clearer. 1. Clarifying the Transformation Law You’re absolutely correct that the transformation law discussed around 3:40 could be misinterpreted as conflating different points. Thank you for pointing this out-it’s an important nuance that I should have addressed more explicitly. The key assumption is that as 𝜆→0, the two points 𝑃 and 𝑄 (which are initially distinct) merge into the same point in the limit. This assumption underpins the idea that we can compare the transformed quantities at 𝑃 and 𝑄 because, in the infinitesimal limit, the difference between these points vanishes. Without this clarification, the logic could appear inconsistent, as you have rightly noted. I'll aim to be more explicit about this in future discussions. To make it rigorous: - At finite separation, 𝑃 and 𝑄 are distinct points, so their respective transformations need to be treated independently. - In the limit 𝜆→0, 𝑄 approaches 𝑃, and this allows us to meaningfully compare the transformations at these two points under the flow generated by the vector field. I’ll take your suggestion to explicitly mention this limit in future videos to avoid confusion. 2. The Role of the Lie Derivative I completely agree with you that the Lie derivative provides a rigorous definition of this concept. Starting with the Lie derivative allows us to bypass any ambiguity and rigorously define how vectors or tensor fields behave under the flow of a vector field 𝑋. For example: 𝐿_𝑋 𝑇 = lim_(𝜆→0) [Φ_(−𝜆)^∗𝑇 − 𝑇]/𝜆 , where Φ_(−𝜆)^∗ is the pullback associated with the flow of 𝑋. This definition mathematically formalizes the infinitesimal change of the tensor 𝑇 along the flow of 𝑋, resolving any issues with conflating points. 3. Hand-Wave Arguments as Interpretation I also agree with your perspective on hand-wave arguments-they’re best viewed as intuitive interpretations of the Lie derivative rather than rigorous derivations. The transport-based arguments are useful for building intuition about how objects like vectors or tensor fields "change" along a flow, but they shouldn’t replace the formal definition. For those who are more mathematically inclined, starting with the Lie derivative's rigorous definition is the clearest approach. From there, one can return to the approximate, geometric interpretations (like visualizing transporting vectors or forms along flows) to build a more intuitive understanding. 4. Improving the Explanation Thank you for emphasizing the need to fill in the details behind the hand-wave arguments. I’ll aim to strike a better balance between mathematical rigor and intuition in future explanations. For viewers looking for a deeper dive, I’d recommend consulting a textbook or resource that carefully develops the Lie derivative, such as Frankel’s The Geometry of Physics or Schutz’s Geometrical Methods of Mathematical Physics. Thank you again for your thoughtful comment and feedback-it’s incredibly helpful in improving my explanations for this material. I hope this response addresses your concerns and provides a clearer picture of the transformation law and its connection to the Lie derivative!

@@TensorCalculusRobertDavie Thank you so much. I can understand the formal definition at 6:20, but it leaves me uneasy. The numerator of the limit seems like it is the difference between the components of the same vector v, expressed in different coordinate systems, the primed and unprimed. Since these components are real numbers, they can be subtracted, but it still feels a little odd to suppress their associated basis vectors, treat them as raw numbers and subtract them. I understand that in an infinitesimal limit the points coincide and the bases may be the same, so perhaps this isn't so odd.? The definition ‘𝐿_𝑋 𝑇 = lim_(𝜆→0) [Φ_(−𝜆)^∗𝑇 − 𝑇]/𝜆 , where Φ_(−𝜆)^∗ is the pullback associated with the flow of X’ also makes great sense. However, for the 'physical 'argument I am struggling with the intuition at 5:50, starting with the line ‘Now we can compare...’.( Physical intuition is a major weakness for me). It seems obvious that v^i(Q) is the vector v evaluated at Q, but why is v‘ ^i(Q) a valid expression of the vector v at point P? I have a vague mental picture of somehow ‘sliding the the coordinate system underneath the point ’ rather that sliding the point along the coordinate line, and that these motions might be somehow equivalent...but it is still very unclear to me. Many thanks if you can provide the insight. BTW: I think it is rare that you put such care and effort into producing the highest quality lessons possible. It’s made me wonder a bit about who you are and what has inspired you. I’d like to find the spring of your idealism and take a sip of that too.

@@johnwarren8032 Thank you for your kind words and your thoughtful question! I’m glad you’re engaging so deeply with the material-it’s clear you’re thinking about both the formal definition and the intuition behind it, which is a great approach. Let me address each of your points in turn. 1. The Formal Definition (6:20): Subtracting the Components You’re absolutely right that the numerator of the Lie derivative formula involves subtracting the components of the vector 𝑣, expressed in different coordinate systems. This might feel strange at first because we’re treating the components as "raw numbers" without explicitly mentioning their associated basis vectors. However, there’s a reason this works in the infinitesimal limit: - When 𝜆 → 0 , the flow map Φ_−𝜆 infinitesimally shifts the vector field 𝑣 from one point 𝑄 back to the neighboring point 𝑃. In this process, the pullback Φ_−𝜆^∗ 𝑣 ensures that 𝑣^𝑖(𝑄) is re-expressed in the same basis at 𝑃. - The subtraction Φ_−𝜆^∗ 𝑣 − 𝑣 is valid because the pullback aligns the vectors into the same tangent space (at 𝑃), so their components can be directly compared. Hence, while the basis vectors are "suppressed" in the notation, their presence is implicit in the pullback operation. This ensures the calculation is mathematically sound. 2. The Intuition at 5:50: Why 𝑣′^𝑖(𝑄) Represents 𝑣 at 𝑃 Your struggle with the physical intuition is very common, and I appreciate your honesty in expressing it! Let’s unpack this step by step: - The key idea is that the flow Φ_−𝜆 moves the coordinates at 𝑄 back to 𝑃. When you evaluate 𝑣 at 𝑄 but express its components in the primed coordinates (which are tied to the flow), 𝑣′^𝑖(𝑄) gives you a "transported" version of 𝑣 as if it were at 𝑃, described in the primed basis. - Your mental image of "sliding the coordinate system underneath the point" is actually quite close to the mark! The flow Φ_−𝜆 effectively shifts the coordinate grid, transforming the vector 𝑣 so that its expression in the primed coordinates reflects how it "would look" if the flow had mapped 𝑄 to 𝑃. Think of the primed coordinates as a new reference frame that travels along the flow. The components 𝑣′^𝑖(𝑄) then capture how the vector 𝑣 aligns with this moving reference frame, even as it’s evaluated at 𝑄. 3. Connecting the Two Perspectives The formal definition and the intuition are connected through the pullback operation Φ_−𝜆^∗ . This map ensures that vectors at 𝑄 are transformed into the tangent space at 𝑃 in a way that respects the geometry of the flow. In essence: - The pullback aligns the tangent spaces so that subtraction is meaningful. - The intuition builds on this alignment by interpreting the transformed components in the primed coordinate system. Your observation that "in the infinitesimal limit the points coincide and the bases may be the same" captures this alignment beautifully. At the core, the Lie derivative compares how the vector field 𝑣 changes relative to the flow generated by 𝑋. 4. A Note on Inspiration Thank you so much for your kind words about the lessons! What inspires me is the opportunity to share the elegance of mathematics and physics in a way that makes these concepts accessible to everyone, regardless of their starting point. Your curiosity and thoughtful engagement are a big part of what makes this work rewarding-it’s a shared journey of exploration and understanding. I hope this clarifies the intuition and formal aspects for you! If you’re still unsure about any part of this, feel free to ask-I’m happy to elaborate further. Thanks again for your thoughtful questions and your support!

Thank you and Merry Christmas to you and your family. Thank you for your earlier comments on one of the videos from push forward of vectors on manifolds part five. I replaced that video with two that were vastly better. Your comments made a real difference.

Thank you so much for your thoughtful feedback and for taking the time to engage with the video. I really appreciate your observations, and I’ll do my best to address your points. 1. Clarification of the Steps at 9:30 You are absolutely right that the explanation at 9:30 combines two conceptual steps that deserve to be treated separately for clarity. Let me outline them explicitly: Step 1: Pullback and the equivalence of integrals This step relies on the property of the pullback, which ensures that the integral of a differential form 𝜔 over a manifold 𝑁 is equal to the integral of the pullback 𝑓^∗(𝜔) over the manifold 𝑀: ∫_𝑁 𝜔 = ∫_𝑀 𝑓^∗(𝜔). The pullback 𝑓^∗(𝜔) is expressed in terms of the pullback of the components of 𝜔, and the orientation-preserving map 𝑓 ensures that the geometric meaning of the integral is preserved. Step 2: Expression of 𝑓^∗(𝜔) in local coordinates Once 𝑓^∗(𝜔) is written in terms of local coordinates on 𝑀, it involves a wedge product like 𝑑𝑢∧𝑑𝑣. At this stage, we can invoke the definition of the integral of a differential form over a Euclidean domain to evaluate it as an iterated integral: ∫_𝑀 𝑓^∗(𝜔) = ∫_𝑀 𝑔(𝑢,𝑣) 𝑑𝑢∧𝑑𝑣 = ∫_𝑢1 ^𝑢2 ∫_𝑣1 ^𝑣2 𝑔(𝑢,𝑣) 𝑑𝑢 𝑑𝑣. This connection between the abstract formalism of differential forms and the familiar iterated integral makes it possible to perform explicit computations. I appreciate your suggestion to separate these steps for clarity. I will make an effort to present this process in a more step-by-step fashion in future videos. 2. Typo at 15:45 Thank you for catching this! You’re absolutely correct that ∫_𝑁 𝑧 𝑑𝑥 𝑑𝑦 is not properly defined in the context of integration on a general manifold 𝑁, as 𝑑𝑥 𝑑𝑦 is not a 2-form but rather a product of differentials from traditional calculus notation. The correct notation should indeed be ∫_𝑁 𝑧 𝑑𝑥∧𝑑𝑦, which represents the integral of the 2-form 𝑧 𝑑𝑥∧𝑑𝑦 over the manifold 𝑁. The wedge product 𝑑𝑥∧𝑑𝑦 ensures that the integral is well-defined on a general manifold, where the geometry is more abstract than a Euclidean domain. Without this correction, the expression could be misinterpreted as attempting to apply iterated integrals directly to a manifold 𝑁, which is not always possible. I’ll update my notes and future explanations to clarify this point and ensure that the correct notation is consistently used to avoid any confusion. Apology Not Necessary! There’s absolutely no need to apologize-you’ve raised excellent points, and your observations are spot-on. Discussions like these are incredibly valuable for improving the clarity and rigor of my explanations. Thank you for helping me refine my presentation, and please don’t hesitate to share any further thoughts or questions!

Thanks for that feedback. Cheers.

Hello Daniel and thank you for spotting that. You are quite right. The basis 1-form dz is not not supposed to be there as the question was dealing with the integral of a 2-form over a surface. It should have read: ω=f(x,y,z)dx∧dy. Thanks again for spotting that.

@@TensorCalculusRobertDavie Thank you for this wonderful channel that has taught me so much!

Thank you for that.

You're welcome. I use Paint 2D, PPT and Mathematica.

Hello Daniel and thank you for that! It should be fixed now.

Thank you! If you mean professor as teacher, then yes to that.

Thank you so much for your kind words and detailed feedback! I really appreciate your observations, and I’m glad the explanation was helpful in filling some gaps. Let me address your points one by one: 1. Clarifying the Role of the Vector Field 𝑋 You make an excellent point about potential confusion when introducing 𝑋 as a "vector field" in the general (𝑥,𝑦) coordinates without explicitly tying it to the curve 𝛾(𝑡) until later. This is something I could definitely improve in future videos. To clarify, 𝑋 is indeed just a single vector at the specific point on the curve 𝛾(𝑡) in this example. Although it’s written in terms of a general coordinate basis (∂/∂𝑥, ∂/∂𝑦), it represents the same object that’s derived from the curve via 𝛾′(𝑡). Your suggestion to explicitly connect 𝑋 to 𝛾′(𝑡) earlier in the discussion is very helpful, and I can see how that would prevent the impression of multiple independent vectors or an unrelated field. Thank you for pointing this out! 2. Detailed Connection Between the Two Expressions for 𝑋(𝑓) Great question about the relationship between the two expressions for 𝑋(𝑓)! Here’s the breakdown: From the curve: - We start with a curve 𝛾(𝑡) = (𝑥(𝑡),𝑦(𝑡)) parametrized by 𝑡. - The tangent vector to this curve is 𝛾′(𝑡) = (𝑑𝑥/𝑑𝑡, 𝑑𝑦/𝑑𝑡). - If 𝑓 is a function defined on the manifold, the rate of change of 𝑓 along the curve is given by: 𝑋(𝑓) = 𝑑𝑓/𝑑𝑡 = ∂𝑓/∂𝑥 𝑑𝑥/𝑑𝑡 + ∂𝑓/∂𝑦 𝑑𝑦/𝑑𝑡 . - This 𝑋(𝑓) captures the derivative of 𝑓 in the direction of 𝛾′(𝑡). General formula for 𝑋(𝑓): - A vector field 𝑋 in coordinates is expressed as: 𝑋 = 𝑋^𝑥 ∂/∂𝑥 + 𝑋^𝑦 ∂/∂𝑦. - Acting on a function 𝑓, 𝑋 evaluates as: 𝑋(𝑓) = 𝑋^𝑥 ∂𝑓/∂𝑥 + 𝑋^𝑦 ∂𝑓/∂𝑦. - Comparing this with the previous expression, we identify: 𝑋^𝑥 = 𝑑𝑥/𝑑𝑡 , 𝑋^𝑦 = 𝑑𝑦/𝑑𝑡. Bringing it together: - The vector 𝛾′(𝑡) defines the components 𝑋^𝑥 and 𝑋^𝑦 in the coordinate basis (∂/∂𝑥, ∂/∂𝑦). - Thus, 𝑋 is expressed as: 𝑋 = 𝑑𝑥/𝑑𝑡 ∂/∂𝑥 + 𝑑𝑦/𝑑𝑡 ∂/∂𝑦. The connection lies in the fact that both formulations describe the same vector: 𝛾′(𝑡) gives its components geometrically, while 𝑋(𝑓) expresses its action as a differential operator. 3. Timestamps for Details At 17:30, I describe 𝑋(𝑓) = 𝛾′(𝑡)⋅∇𝑓, which connects the curve's derivative 𝛾′(𝑡) to the function's gradient ∇𝑓. At 21:30, I introduce the explicit coordinate representation 𝑋(𝑓) = 𝑋^𝑥 ∂𝑓/∂𝑥 + 𝑋^𝑦 ∂𝑓/∂𝑦. The key connection between the two is identifying 𝑋^𝑥 and 𝑋^𝑦 as the components of 𝛾′(𝑡) in the coordinate basis. If any of this was unclear in the video, feel free to let me know, and I’d be happy to provide additional explanations. Thank you so much for your kind holiday wishes! I genuinely enjoy sharing these ideas, and your thoughtful engagement is the best reward. I don’t have a Patreon or tip system, but your gratitude and feedback mean the world to me. Happy holidays to you as well, and I look forward to hearing from you again!

You're welcome. Thank you for your comments. They are very helpful for this channel.

Thank you for that feedback. Glad you liked it.

Please see my responses to that question.

Thank you for saying that!

I know what you mean. Good intentions and all that ......

Thank you for that. Your comment gets to the heart of what I am trying to achieve on this channel.

Thank you for your question because it is a very good one, since it is not obvious why that choice was made! I chose not explain it because I thought it would distract from the rest of the content in the video. At 17:41 in the video, I introduce 𝜙 = Δ𝜙/2 as part of the geometric or analytical setup. Here's the reasoning: 1. Context: In the video, we are likely examining an angular change (or interval) Δ𝜙, which represents the total angular span or difference. To perform certain calculations or analyses, it's often useful to consider the midpoint or symmetry of the angular interval. In this case, 𝜙 is chosen to represent the "half-angle" of the interval, i.e., half of Δ𝜙. 2. Why 𝜙 = Δ𝜙/2? By taking 𝜙 = Δ𝜙/2 , we are essentially centering the angular interval symmetrically around a midpoint. This is especially useful in scenarios where: - The angular interval Δ𝜙 corresponds to a small rotation or change in angle. - We are working with trigonometric functions (like sine or cosine), where halving the angle simplifies certain expressions or calculations (e.g., double-angle formulas). - The geometry or physics of the problem is naturally symmetric, so using the half-angle simplifies visualization and computation. 3. How this is applied: Suppose Δ𝜙 represents the total angular range between two points. Taking 𝜙 = Δ𝜙/2 is a common technique to focus on a representative angle for symmetric contributions or to split the total range into equal halves. For example: - In many integration problems over angular variables, symmetry arguments allow us to calculate over half the interval and then double the result. - In some derivations, the half-angle substitution reduces the complexity of the equations. If you’d like, I can explain this in greater detail or clarify how it fits into the specific context of the video. Let me know!

@TensorCalculusRobertDavie i understand half of what you written here cuz it seems heavy assumptions. So it would be quite helpful if you explain this both geometrically (using this figure and angular representation of the day diagram)and analytically in other video. Thank you for your kind reply

@@Scientificirfann Thank you for your follow-up question! I completely understand that some of the concepts may initially seem abstract or involve heavy assumptions. In physics, making assumptions and approximations is often not just useful-it’s essential. Let me explain why this approach is valid and how it helps build both intuition and analytical solutions. Why Assumptions Matter In real-world systems, the complexity of phenomena can make direct solutions practically impossible. Assumptions help simplify these systems by focusing on the essential features that influence the problem while ignoring less critical details. Examples of Common Assumptions in Physics 1. Point Mass Approximation: When studying planetary motion, we often treat planets as point masses even though they are complex, distributed objects. 2. Frictionless Surfaces: In introductory mechanics, assuming no friction allows us to isolate and study fundamental forces like gravity. 3. Linear Approximations: For small oscillations around an equilibrium, non-linear systems can often be approximated by linear equations that are much easier to solve. Approximations as a Tool for Progress 1. First-Order Understanding: Approximations give us a way to make predictions that are "good enough" to match experimental data within acceptable margins of error. 2. Building Intuition: By starting simple and introducing approximations, we gain insights that guide our understanding of more complex systems. 3. Iterative Refinement: After solving the simplified version, more complex effects can be gradually introduced to improve accuracy. In my previous explanation, certain assumptions and approximations have been implied to streamline the discussion. This method is common when: Geometrical representations help visualize abstract concepts. Analytical techniques simplify expressions to focus on fundamental relationships. Thank you for your thoughtful engagement. In physics, balancing assumptions with detailed analysis is a skill developed over time, and questions like yours are key to mastering this balance.

Thank you so much for your thoughtful question and for pointing out the parallels between the two approaches-tensor analysis (as discussed in Synge and Schild) and the differential forms perspective. I appreciate the opportunity to explore this connection, as it highlights the richness of geometric and algebraic structures in differential geometry and physics. Let me address your question in parts: 1. Volume Elements in Tensor Analysis and Differential Forms - In tensor analysis, as Synge and Schild describe, the concept of volume is tied to the "extension" of an 𝑀-cell, which arises from the wedge product of tangent vectors spanning the cell. This product defines a contravariant object in the tangent space, reflecting its dependence on basis vectors. The transformation properties of this volume element involve compensating factors from the metric tensor (weight +1) and the Jacobian determinant of the coordinate transformation (weight −1), ensuring invariance of the volume under coordinate transformations. - In the differential forms framework, the volume element is defined as a top-degree differential form-essentially an antisymmetric covariant tensor. This form lives in the cotangent space and transforms in such a way that it "absorbs" the effects of coordinate transformations, producing an invariant volume element suitable for integration. When a metric is introduced, it allows us to explicitly relate differential forms to objects in the tangent space, as the metric provides a natural isomorphism between tangent and cotangent spaces. 2. Behavior Under Coordinate Transformations Both approaches guarantee invariance of the volume element but conceptualize it differently: - In tensor analysis, the invariance arises as the product of two separate effects: the metric tensor (associated with the geometry) and the Jacobian determinant (associated with the coordinate transformation). - In the differential forms approach, the Jacobian determinant appears intrinsically in the transformation properties of the differential form itself, eliminating the need to separately account for it. The metric, if present, may define the volume form in terms of the Hodge star operator, which maps to the top-degree form. 3. Covariant vs. Contravariant Nature You're absolutely right about the distinction: - The extension of an 𝑀-cell in tensor analysis is a contravariant object, as it involves tangent vectors. - The volume element as a differential form is covariant, as it belongs to the cotangent space. However, these are deeply connected by duality. The metric allows us to move between the tangent and cotangent spaces, bridging the contravariant and covariant perspectives. In particular, the contravariant volume element in tensor analysis can be pulled into the cotangent space via the metric to correspond directly to the volume form. 4. Connections Between Differential Forms and Tensors The differential forms approach can be viewed as a reformulation of many ideas from tensor analysis, emphasizing intrinsic geometric properties rather than component-level expressions. Here are some key parallels: Divergence: The divergence of a vector field (in tensor language) corresponds to the exterior derivative of the (𝑛−1)-form dual to the vector field in differential forms. Curl: The curl of a vector field in 3D can be expressed using the Hodge star operator, mapping a 1-form to a 2-form. Integration: In tensor analysis, integration often involves the determinant of the metric to define the correct volume measure. In differential forms, the volume form naturally incorporates this determinant through the Hodge star. 5. Insights on the Two Perspectives - Tensor analysis is often more direct when describing physics in terms of vectors and tensors in the tangent space. It provides an intuitive "building-block" approach for many applications. - Differential forms, on the other hand, excel in abstracting intrinsic geometric properties and are particularly powerful in generalizing ideas to manifolds of arbitrary dimension and topology. Ultimately, the choice between these formulations depends on the problem at hand and personal preference. Both are mathematically equivalent in most cases and can often be translated into one another using the metric as a bridge.

In Synge and Schild's book Tensor Calculus, an M-cell refers to a geometric construct used to define the concept of volume in a manifold. This is tied to their approach to extending classical notions of volume in 𝑅^𝑛 to more general curved spaces. Key Ideas from Synge and Schild: 1. M-cell as a Parallelepiped: An 𝑀-cell in this context is an 𝑛-dimensional parallelepiped spanned by 𝑛 linearly independent tangent vectors in the tangent space of a manifold at a given point. These vectors, often denoted as 𝑣_1, 𝑣_2, …, 𝑣_𝑛 , define a geometric volume element in the tangent space. 2. Extension of an M-cell: The extension of the 𝑀-cell refers to the 𝑛-dimensional volume spanned by these vectors, calculated using the determinant of their components with respect to a chosen coordinate system. Specifically, the volume is given by the determinant of the matrix formed by the components of the vectors under consideration: Volume = sqrt[det(𝑔_𝑖𝑗), where 𝑔_𝑖𝑗 is the metric tensor evaluated at the point in question. 3. Coordinate Transformation and Tensor Weights: Synge and Schild emphasize how the 𝑀-cell transforms under changes of coordinates: The coordinate transformation affects the components of the vectors spanning the 𝑀-cell, leading to a scaling factor (the Jacobian determinant) that reflects how the volume element is stretched or compressed. The metric tensor provides an additional weight, reflecting the curvature or geometry of the manifold. 4. Invariant Volume Element: The volume element of an 𝑀-cell is ultimately an invariant quantity: 𝑑𝑉 = sqrt[det(𝑔_𝑖𝑗)] 𝑑𝑥^1 ∧ 𝑑𝑥^2 ∧ ⋯ ∧ 𝑑𝑥^𝑛 , where sqrt[det(𝑔_𝑖𝑗)] accounts for the metric's influence on the 𝑀-cell's size and the wedge product encodes orientation. Comparison with Differential Forms: Synge and Schild's 𝑀-cell approach aligns with the concept of differential forms in many ways: Differential forms (e.g., an 𝑛-form) generalize the idea of oriented volume elements. The volume form in differential geometry is 𝜔 = sqrt[det(𝑔_𝑖𝑗)] 𝑑𝑥^1 ∧ 𝑑𝑥^2 ∧ ⋯ ∧ 𝑑𝑥^𝑛 , which is conceptually equivalent to the invariant volume derived from 𝑀-cells. Why Use the Term "M-cell"? The term emphasizes the geometric intuition behind volume: It highlights the use of parallelepipeds spanned by tangent vectors as fundamental building blocks for defining and calculating volumes. In Synge and Schild's terminology, the "M" in M-cell stands for manifold. The concept is rooted in the geometric structure of the manifold and emphasizes the local behavior of volume elements within it. An M-cell specifically refers to a small, 𝑛-dimensional region in the manifold, conceptualized as being spanned by 𝑛 tangent vectors at a point. The term is used to connect the abstract manifold to a more intuitive, geometric representation of volume or space within it. This naming aligns with the broader usage of "M" in the context of manifold-related structures, such as 𝑀-dimensional volumes, 𝑀-dimensional coordinate systems, and so on. The 𝑀-cell connects abstract mathematical constructs like tensors and forms to tangible geometric objects.

​@@TensorCalculusRobertDavie Thank you for the excellent, thorough answer. I would be most interested to see more details of the explicit translations. I understand some of them are subsumed in the translation of the Gibbs notation to the language of forms, but the tensor language has more to it than that and there is more to do. Perhaps a subject for a future video? Thanks again!

@@TensorCalculusRobertDavie Thank you for the excellent, thorough answer. I find this part of Synge and Schild to be a bit dense and I wish they had spent a bit more time and space explaining these intricacies. Very helpful.

Ahh! I see now! It is correct, because of the symmetry of R under the exchange of the first and the second pairs of indices! I get it..

Yes, it is!

Please see my response to the previous video you refer to.