answer:The Hopf map is a mapping from the 3-sphere S^3 to the 2-sphere S^2, which can be represented using quaternions. Quaternions are a number system that extends complex numbers, with three imaginary units i, j, k such that i^2 = j^2 = k^2 = ijk = -1. A quaternion q can be written as q = a + bi + cj + dk where a, b, c, d are real numbers and a is the real part, while bi + cj + dk is the imaginary part. The conjugate of a quaternion is given by bar{q} = a - bi - cj - dk, and its norm is |q| = sqrt{qbar{q}} = sqrt{a^2 + b^2 + c^2 + d^2}. The 3-sphere S^3 can be represented as the set of all quaternions with norm 1, i.e., S^3 = {q in mathbb{H} : |q| = 1}. The Hopf map h: S^3 to S^2 can be defined using quaternion multiplication as follows: h(q) = qibar{q} = (bi - cj + dk) / sqrt{b^2 + c^2 + d^2} Note that h(q) is a pure imaginary quaternion, i.e., it has no real part. The norm of h(q) is always 1, so h(q) in S^2. Now, let's understand the Hopf fibration intuitively. Consider a point p in S^2 and its preimage h^{-1}(p) in S^3. This preimage is a circle (known as a fiber) that lies in S^3. To see this, note that if q in h^{-1}(p), then qibar{q} = p, which implies qi = pq. Squaring both sides, we get (qi)^2 = p^2q^2, or qi^2q^{-1} = -p^2. Since p in S^2, we have p^2 = -1, so qi^2q^{-1} = q(-q^{-1}q)q^{-1} = -1. This means that qi is a unit imaginary quaternion, and since qi = pq, we have qi and q(-i) = -qi both lying in h^{-1}(p). Thus, h^{-1}(p) forms a circle in S^3. To visualize the Hopf fibration, we can think of S^3 as the surface of a 4-dimensional sphere, while S^2 as the surface of a 3-dimensional sphere. For each point p in S^2, consider the set of all lines passing through the origin that intersect S^3 at a point in the preimage h^{-1}(p). These lines form a 2-dimensional plane in mathbb{R}^4 (known as a Hopf circle), and the set of all such planes forms a solid torus T in mathbb{R}^4. The Hopf fibration can then be visualized as the projection of S^3 onto this solid torus, with each point in S^3 mapping to a circle in T. Note that the Hopf fibration is not a trivial fibration, i.e., it cannot be realized as a product space of S^1 and S^2. This has important consequences in the study of topology and quantum physics.,The Hopf map is a continuous function from the 3-sphere S^3 to the 2-sphere S^2. To define the Hopf map using quaternions, let's first represent points in S^3 as unit quaternions. A quaternion q can be written as: q = a + bi + cj + dk where a, b, c, and d are real numbers, and i, j, and k are imaginary units with the property i^2 = j^2 = k^2 = ijk = -1. A unit quaternion has the property: a^2 + b^2 + c^2 + d^2 = 1 Now, let's define the Hopf map h: S^3 → S^2. Given a unit quaternion q in S^3, the Hopf map h(q) is a point in S^2, which can be represented as a complex number. The formula for the Hopf map is: h(q) = q * i * q^(-1) where q^(-1) is the inverse of q, which can be computed as: q^(-1) = a - bi - cj - dk Now, let's compute h(q): h(q) = (a + bi + cj + dk) * i * (a - bi - cj - dk) = (a + bi + cj + dk) * (ai - b - ci - di) = (a^2 - ab - ac - ad)i + (ab - b^2 - bc - bd) + (ac - bc - c^2 - cd)j + (ad - bd - cd - d^2)k Since h(q) is a point in S^2, we can represent it as a complex number: h(q) = (ab - b^2 - bc - bd) + (a^2 - ab - ac - ad)i Now, let's sketch the Hopf fibration. The Hopf fibration can be visualized as a collection of circles (fibers) in S^3 that are mapped to points in S^2 by the Hopf map. Each fiber is a circle that lies on a plane in 4-dimensional space, and each point on the circle corresponds to a unique point in S^2. To visualize the Hopf fibration, imagine a 2-sphere S^2 in 3-dimensional space. Each point on S^2 corresponds to a circle in S^3. The circles are linked together in a way that resembles a torus, with each circle passing through the "hole" of the torus. The Hopf map maps each circle in S^3 to a single point in S^2. In summary, the Hopf map is a continuous function from the 3-sphere S^3 to the 2-sphere S^2, defined using quaternions as the representation of points in S^3. The formula for the Hopf map is h(q) = q * i * q^(-1), and the Hopf fibration can be visualized as a collection of linked circles in S^3 that are mapped to points in S^2.,The Hopf map, denoted by h: S^3 → S^2, is a continuous map defined using quaternions. Let q = (q0, q1, q2, q3) be a point in S^3, where the distance from the origin is 1: q0^2 + q1^2 + q2^2 + q3^2 = 1. The Hopf map h(q) is given by the formula: h(q) = (2(q0q2 + q1q3), 2(q1q2 - q0q3), q0^2 + q1^2 - q2^2 - q3^2) Now, to sketch the Hopf fibration, we need to understand the relationship between the preimages of points in S^2 and the points in S^3. The Hopf fibration can be visualized as a collection of circles (called fibres) in S^3, each of which maps to a point in S^2 under the Hopf map. The fibres can be represented as the collection of all points (q0, q1, q2, q3) in S^3 that satisfy q0^2 + q1^2 = q2^2 + q3^2. These fibres are parallel to the quaternion imaginary plane, and they are evenly distributed in S^3. Each fibre maps to a meridian on S^2, which are the lines of constant longitude. To visualize the Hopf fibration, you can imagine S^3 as a 3-dimensional sphere surrounding another 2-dimensional sphere, S^2 (such as the Earth). Each fibre, represented by a circle in S^3, is attached to the 3-sphere like the thread of a screw, and each circle wraps around the 3-sphere and crosses through the equator of the 3-sphere to reach the "Earth." When you follow a fibre from the north pole of S^3 to the south pole, it becomes a great circle on S^2, which corresponds to a meridian on the Earth. In summary, the Hopf map h: S^3 → S^2 is defined by the formula h(q) = (2(q0q2 + q1q3), 2(q1q2 -

answer:The Hopf map is a map from the 3-sphere (a 3-dimensional manifold) to the 2-sphere (a 2-dimensional manifold). In this case, we're not actually working with the full 3-sphere or 2-sphere, but rather certain submanifolds of them. Specifically, we're considering the manifolds M and N that result from revolving circles about the y-axis. The given map h: M → N is defined by h(x,y,z) = x^2 - y^2, 2xy. To understand what this map does, it's helpful to rewrite it in terms of spherical coordinates. Recall that the relationship between Cartesian and spherical coordinates in 3 dimensions is given by: x = r sin(θ) cos(φ) y = r sin(θ) sin(φ) z = r cos(θ) where r is the radial distance, θ is the polar angle (measured from the positive z-axis), and φ is the azimuthal angle (measured in the xy-plane from the positive x-axis). Since M is obtained by revolving the circle x^2 + (y-1)^2 = 1 about the y-axis, we can parameterize M using the angles θ and φ as follows: x = sin(θ) cos(φ) y = sin(θ) sin(φ) + 1 z = cos(θ) Note that r = 1 here since M is a unit circle in the plane y = 1. Now let's compute the image of h(x,y,z) for a point on M: h(x,y,z) = h(sin(θ) cos(φ), sin(θ) sin(φ) + 1, cos(θ)) To compute this, we first need to find x^2 - y^2 and 2xy. We have: x^2 - y^2 = sin^2(θ) cos^2(φ) - (sin(θ) sin(φ) + 1)^2 = sin^2(θ) cos^2(φ) - sin^2(θ) sin^2(φ) - 2sin(θ) sin(φ) - 1 and 2xy = 2sin(θ) cos(φ)(sin(θ) sin(φ) + 1). Now, note that N is obtained by revolving the circle x^2 + y^2 = 1 about the y-axis, so we can parameterize N using a single angle φ as follows: x = cos(φ) y = sin(φ) The Hopf map h should take values in N, so we need to express h(x,y,z) in terms of the angular coordinate φ on N. To do this, we'll eliminate θ from our expressions for x^2 - y^2 and 2xy. We have: (x^2 - y^2)^2 + (2xy)^2 = (sin^2(θ) cos^2(φ) - sin^2(θ) sin^2(φ) - 2sin(θ) sin(φ) - 1)^2 + (2sin(θ) cos(φ)(sin(θ) sin(φ) + 1))^2 = sin^4(θ) cos^4(φ) - 2sin^4(θ) cos^2(φ) sin^2(φ) + sin^4(θ) sin^4(φ) - 4sin^3(θ) cos(φ) sin(φ) + 4sin^2(θ) cos^2(φ) To proceed further, note that the point x,y,z on M lies on the unit sphere, so x^2 + y^2 + z^2 = 1, which gives us: sin^2(θ) cos^2(φ) + sin^2(θ) sin^2(φ) + cos^2(θ) = 1 sin^2(θ) (cos^2(φ) + sin^2(φ)) = 1 - cos^2(θ) sin^2(θ) = 1 - cos^2(θ) Substituting this back into our expression for h(x,y,z), we get: (x^2 - y^2)^2 + (2xy)^2 = (1 - cos^2(θ)) cos^4(φ) - 2(1 - cos^2(θ)) cos^2(φ) sin^2(φ) + (1 - cos^2(θ)) sin^4(φ) - 4(1 - cos^2(θ)) cos(φ) sin(φ) + 4(1 - cos^2(θ)) cos^2(φ) = cos^4(φ) - 2cos^2(φ) sin^2(φ) + sin^4(φ) - 4cos(φ) sin(φ) + 4cos^2(φ) Now we can express this in terms of the angle φ on N by noting that: cos(φ) = x sin(φ) = y Thus, our final expression for the Hopf map h: M → N is: h(x,y,z) = x^2 - y^2, 2xy = cos^2(φ) - sin^2(φ), 2cos(φ) sin(φ). Since cos^2(φ) + sin^2(φ) = 1, we can also write this as: h(x,y,z) = cos(2φ), sin(2φ). This is indeed a map from M to N, as desired.,The problem statement seems to have a slight inconsistency, as the Hopf map is typically defined as a map from a 3-manifold to a 2-manifold, while in this case, we have two 3-manifolds. It would be more suitable to ask for the Hopf fibration instead of the Hopf map since it relates 3-manifolds and 2-manifolds. Nonetheless, let's assume that you meant the Hopf fibration between a sphere and a 4-sphere. In that case, we can still try to determine if there is a relationship between the given expressions and the Hopf fibration. The Hopf fibration is a map H: S^3 → S^2 that interweaves the circles in S^3 into a nontrivial way. Here, S^3 represents a three-dimensional sphere, and S^2 represents a two-dimensional sphere. The traditional map is defined as H(z, w) = (2zw), (z^2 + w^2), where z and w are complex numbers. However, your h: M → N function does not seem to fit this description. The domain and range of your function (M and N) are both 3-dimensional, and the functional form x^2− y^2, 2xy doesn't match the traditional Hopf fibration formula. In conclusion, your given function h: M → N seems unrelated to the Hopf fibration as described for spheres. If the alignment was just a misunderstanding, please provide more information or context for a better and more relevant response.,To determine the Hopf map h: M → N given by h(x, y, z) = x^2 - y^2, 2xy for x, y, z in M, we first need to parameterize the points on M and N. Let's parameterize M using cylindrical coordinates. Let r = 1, and let θ be the angle between the positive x-axis and the line connecting the origin to the point (x, y) in the xy-plane. Then, we can parameterize M as follows: x = cos(θ) y = 1 + sin(θ) z = z Now, let's parameterize N using polar coordinates. Let R = 1, and let φ be the angle between the positive x-axis and the line connecting the origin to the point (x, y) in the xy-plane. Then, we can parameterize N as follows: x = cos(φ) y = sin(φ) Now, we can express the Hopf map h in terms of θ and z: h(cos(θ), 1 + sin(θ), z) = cos^2(θ) - (1 + sin(θ))^2, 2cos(θ)(1 + sin(θ)) To find the corresponding point on N, we need to find the angle φ such that: cos^2(φ) = cos^2(θ) - (1 + sin(θ))^2 sin^2(φ) = 2cos(θ)(1 + sin(θ)) Adding these two equations, we get: cos^2(φ) + sin^2(φ) = cos^2(θ) - (1 + sin(θ))^2 + 2cos(θ)(1 + sin(θ)) Since cos^2(φ) + sin^2(φ) = 1, we have: 1 = cos^2(θ) - (1 + sin(θ))^2 + 2cos(θ)(1 + sin(θ)) Now, we can solve for φ in terms of θ: cos(φ) = sqrt(cos^2(θ) - (1 + sin(θ))^2 + 2cos(θ)(1 + sin(θ))) sin(φ) = sqrt(2cos(θ)(1 + sin(θ))) Thus, the Hopf map h: M → N is given by: h(cos(θ), 1 + sin(θ), z) = cos(φ), sin(φ) where φ is determined by the equations above.

answer:To find the Killing vectors, we need to solve the Killing equation: ∇_u g = 0 where u is a vector field and ∇ is the covariant derivative associated with the metric g. In local coordinates, this becomes a system of partial differential equations for the components of u. Let u = u^1 partial_x + u^2 partial_y be a vector field on the manifold. Then, the Killing equation can be written as (g_{ij,k} + g_{kj,i} - g_{ik,j})u^j = 0 Here, g_{ij} = begin{pmatrix} 1+x^2 & x x & 1 end{pmatrix} are the components of the metric tensor, and g_{ij,k} = frac{partial g_{ij}}{partial x^k} denote their partial derivatives. Computing the required derivatives, we get: g_{11,1} = 2x, g_{12,1} = 1, g_{22,1} = 0 g_{11,2} = g_{12,2} = g_{22,2} = 0 Now, plugging these into the Killing equation, we obtain (2x u^1 + u^2)_x = 0 u^1_y = 0 Integrating the second equation, we see that u^1 = u^1(x), i.e., it depends only on x. Now, we can rewrite the first equation as 2x (u^1)'_x + u^2_x = 0 This is a first-order linear PDE for u^2 in terms of u^1. Its general solution can be found by the method of characteristics: frac{d u^2}{d x} = -frac{2x}{1+x^2} u^2 This ODE has the general solution u^2(x,y) = C(y) (1+x^2)^{-1} for an arbitrary function C(y). Now, plugging this back into the first equation, we obtain u^1(x,y) = -int frac{C'(y)}{1+x^2} dy + D(x) for another arbitrary function D(x). Thus, the general form of a Killing vector is given by u = -left(int frac{C'(y)}{1+x^2} dy + D(x)right) partial_x + C(y) (1+x^2)^{-1} partial_y for arbitrary functions C(y) and D(x).,To find the Killing vector fields of the given Riemannian manifold with the metric tensor g, we need to solve the Killing equation: ∇_μ X_ν + ∇_ν X_μ = 0, where X is the Killing vector field and ∇ denotes the covariant derivative. This equation can be written in terms of the Christoffel symbols Γ^k_ij: X^k (Γ^i_{kj} + Γ^i_{jk}) = 0. First, we need to compute the Christoffel symbols for the given metric tensor. The Christoffel symbols are given by: Γ^k_ij = (1/2) g^(kl) (∂_i g_jl + ∂_j g_il - ∂_l g_ij), where g^(kl) is the inverse of the metric tensor g. For the given metric tensor, the non-zero components of the inverse metric tensor are: g^11 = (1 + x^2 + y^2)^(-1), g^22 = (1 + x^2)^(-1), g^33 = 1. Now, we can compute the non-zero Christoffel symbols for the given metric tensor: Γ^1_11 = (1 + x^2 + y^2) x, Γ^1_12 = Γ^1_21 = x, Γ^1_22 = -(1 + x^2) x, Γ^2_12 = Γ^2_21 = y, Γ^2_22 = (1 + x^2) y, Γ^3_13 = Γ^3_31 = x, Γ^3_23 = Γ^3_32 = y, The other Christoffel symbols are zero. Next, we can write down the Killing equations using the non-zero Christoffel symbols from above. The equations are: X^1 (Γ^1_{11} + Γ^1_{11}) + X^2 (Γ^1_{21},To find the Killing vector fields of the given metric, we need to solve the Killing equation: ∇_μ X_ν + ∇_ν X_μ = 0 where X is the Killing vector field and ∇ is the covariant derivative. The metric tensor g is given by: g = (1 + x^2)dx^2 + 2xdxdy + dy^2 We can write this metric tensor in matrix form as: g = | 1 + x^2, x | | x, 1 | Now, we need to calculate the inverse metric tensor g^μν: g^(-1) = | 1, -x | | -x, 1 + x^2 | Next, we need to calculate the Christoffel symbols Γ^σ_μν, which are given by: Γ^σ_μν = 1/2 * g^(σλ) (∂_μ g_νλ + ∂_ν g_μλ - ∂_λ g_μν) After calculating the Christoffel symbols, we can write the Killing equation in terms of the Christoffel symbols: ∂_μ X_ν + ∂_ν X_μ - Γ^σ_μν X_σ - Γ^σ_νμ X_σ = 0 Now, we need to solve this system of partial differential equations for the components of the Killing vector field X. In general, this can be a difficult task, but in this case, we can make some simplifications. First, let's consider the case where the Killing vector field is a linear combination of the coordinate vector fields ∂/∂x and ∂/∂y: X = A(x,y) ∂/∂x + B(x,y) ∂/∂y Now, let's plug this into the Killing equation and separate it into its components: (1) ∂_x A + ∂_y B - x(∂_x B + ∂_y A) = 0 (2) ∂_x B + ∂_y A - x(∂_x A + ∂_y B) = 0 These are two coupled first-order partial differential equations for the functions A(x,y) and B(x,y). We can solve them by assuming that A and B are separable, i.e., A(x,y) = A_x(x) * A_y(y) and B(x,y) = B_x(x) * B_y(y). Plugging this ansatz into the equations (1) and (2), we get: (1) A_x' * A_y + B_x' * B_y - x(A_x * A_y' + B_x * B_y') = 0 (2) A_x' * B_y + B_x' * A_y - x(A_x * B_y' + B_x * A_y') = 0 Now, we can solve these equations for the functions A_x, A_y, B_x, and B_y. After solving these equations, we can construct the Killing vector fields X by multiplying the solutions for A_x, A_y, B_x, and B_y: X = (A_x(x) * A_y(y)) ∂/∂x + (B_x(x) * B_y(y)) ∂/∂y In this case, the Killing vector fields will be functions of x and y, and they will satisfy the Killing equation for the given metric tensor.

answer:To find the Killing vector fields of a Riemannian metric g, we need to solve the following equation for X: L_X g = 0 where L_X g is the Lie derivative of the metric with respect to the vector field X. This equation is called the Killing equation. In local coordinates, this equation becomes: X^k partial_k g_{ij} + g_{jk} partial_i X^j + g_{ik} partial_j X^k = 0 For our given metric g = dx^2 + dy^2 + 2(x+y)dxdy, we have: g_{11} = 1, g_{12} = g_{21} = x+y, g_{22} = 1 The Killing equation then becomes: X^1 partial_1 g_{ij} + X^2 partial_2 g_{ij} + g_{jk} partial_i X^j + g_{ik} partial_j X^k = 0 Expanding this out, we get: X^1 + (x+y)partial_1 X^1 + partial_1 X^2 + (x+y)partial_2 X^1 + partial_2 X^2 = 0 X^1 + (x+y)partial_1 X^2 + partial_1 X^1 + partial_2 X^1 + (x+y)partial_2 X^2 = 0 Now we equate coefficients of the different terms separately: 1) Coefficients of X^1: partial_1 X^1 + partial_1 X^1 = 0 Rightarrow partial_1 X^1 = 0 2) Coefficients of X^2: partial_2 X^2 + partial_2 X^2 = 0 Rightarrow partial_2 X^2 = 0 3) Coefficients of (x+y)partial_1 X^1: (x+y)partial_1^2 X^1 + (x+y)partial_1^2 X^1 = 0 Since x+y is not identically zero, we can divide by it and get: partial_1^2 X^1 = 0 4) Coefficients of partial_1 X^2: partial_1^2 X^2 + partial_2 partial_1 X^1 = 0 Using our previous results, this simplifies to: partial_1^2 X^2 = 0 5) Coefficients of partial_2 X^1: partial_2^2 X^1 + partial_1 partial_2 X^2 = 0 Using our previous results, this simplifies to: partial_2^2 X^1 = 0 From these equations, we see that the general solution for X=(X^1,X^2) is: X^1 = a_1x + a_2y + a_3 X^2 = b_1x + b_2y + b_3 where a_i, b_i are constants. Now let's verify that these form a Lie algebra under the Lie bracket operation. To do this, we need to compute the Lie bracket X,Y for two Killing vector fields X and Y: X,Y = XY - YX and show that this is also a Killing vector field. In our case, let X=(X^1,X^2) and Y=(Y^1,Y^2) be two Killing vector fields. Then their Lie bracket is: X,Y = (X^1Y^2_x - Y^1X^2_x, X^1Y^2_y - Y^1X^2_y) - (X^2Y^1_x - Y^2X^1_x, X^2Y^1_y - Y^2X^1_y) where subscripts denote partial derivatives. Now plugging in the expressions for X^1, X^2, Y^1, Y^2, we get: X,Y = (a_1b_2 - b_1a_2)(x,y) So the Lie bracket of any two Killing vector fields is again a Killing vector field. This means that the Killing vector fields form a Lie algebra under the Lie bracket operation. In summary, the Killing vector fields of the given Riemannian metric g are: X = (a_1x + a_2y + a_3, b_1x + b_2y + b_3) and they form a Lie algebra under the Lie bracket operation.,To find the Killing vector fields of the given Riemannian metric, we need to solve the Killing equation: nabla_X Y + nabla_Y X = 0 where X and Y are vector fields on mathbb{R}^2, and nabla is the Levi-Civita connection associated with the metric g. Let X = X^1 frac{partial}{partial x} + X^2 frac{partial}{partial y} and Y = Y^1 frac{partial}{partial x} + Y^2 frac{partial}{partial y}. Then the Killing equation can be written as a system of partial differential equations: frac{partial X^1}{partial x} + frac{partial X^2}{partial y} + (x+y)frac{partial Y^1}{partial y} + (x+y)frac{partial Y^2}{partial x} = 0 frac{partial Y^1}{partial x} + frac{partial Y^2}{partial y} + (x+y)frac{partial X^1}{partial y} + (x+y)frac{partial X^2}{partial x} = 0 We can solve this system of PDEs to find the Killing vector fields. After some calculations, we find that the general solution is given by: X = (a + bx) frac{partial}{partial x} + (c + dy) frac{partial}{partial y} Y = (a' + b'x) frac{partial}{partial x} + (c' + d'y) frac{partial}{partial y} where a, b, c, d, a', b', c', d' are constants. The Killing vector fields are linear combinations of the following basis vector fields: K_1 = frac{partial}{partial x} K_2 = frac{partial}{partial y} K_3 = x frac{partial}{partial x} + y frac{partial}{partial y} Now, we need to verify that these Killing vector fields form a Lie algebra under the Lie bracket operation. The Lie bracket of two vector fields X and Y is defined as X, Y = XY - YX. We compute the Lie brackets of the basis Killing vector fields: K_1, K_2 = K_1K_2 - K_2K_1 = 0 K_1, K_3 = K_1K_3 - K_3K_1 = K_1 K_2, K_3 = K_2K_3 - K_3K_2 = K_2 We see that the Lie brackets of the basis Killing vector fields are linear combinations of the basis Killing vector fields themselves. Therefore, the Killing vector fields form a Lie algebra under the Lie bracket operation.,To find the Killing vector fields of the given Riemannian metric, we must first compute the Christoffel symbols Gamma^i_{jk}. The metric tensor g can be written as a matrix: g = begin{pmatrix} 1 + (x+y)^2 & x+y x+y & 1 end{pmatrix} Now, we compute the inverse metric tensor g^{ij}: g^{-1} = frac{1}{(1+(x+y)^2)(1) - (x+y)^2} begin{pmatrix} 1 & -(x+y) -(x+y) & 1+(x+y)^2 end{pmatrix} Next, we compute the Christoffel symbols: Gamma^i_{jk} = frac{1}{2} g^{il} (frac{partial g_{lj}}{partial x^k} + frac{partial g_{lk}}{partial x^j} - frac{partial g_{jk}}{partial x^l}) We need to find the Killing vector fields, which are the vector fields that satisfy the Killing equation: nabla_i X_j + nabla_j X_i = 0 This is equivalent to the following system of equations: frac{partial X_1}{partial x^1} + Gamma^1_{11} X_1 + Gamma^1_{12} X_2 + Gamma^2_{11} X_1 + Gamma^2_{12} X_2 = 0 frac{partial X_1}{partial x^2} + Gamma^1_{21} X_1 + Gamma^1_{22} X_2 + Gamma^2_{21} X_1 + Gamma^2_{22} X_2 = 0 frac{partial X_2}{partial x^1} + Gamma^1_{11} X_2 + Gamma^1_{12} X_1 + Gamma^2_{11} X_2 + Gamma^2_{12} X_1 = 0 frac{partial X_2}{partial x^2} +