Show that for a 1-form \(\alpha\) and for a 2-form \(\omega\), the components of the Lie derivative along \(v \in \mathcal{X}(M)\) are respectively $$ \begin{aligned} \left(\mathcal{L}_{v} \alpha\right)_{a} &=v_{b} \partial_{b} \alpha_{a}+\alpha_{b} \partial_{a} v_{b} \\ \left(\mathcal{L}_{v} \omega\right)_{a b} &=v_{c} \partial_{c} \omega_{a b}+\omega_{c b} \partial_{a} v_{c}+\omega_{a c} \partial_{b} v_{c} \end{aligned} $$ Hence check the key properties of the Lie derivative for 1-forms and 2-forms.

Recall that a 1-form \(\alpha\) is an element of the covectors space and is represented by \(\alpha(\cdot)=\alpha_{a}dx^a\). A 2-form \(\omega\) can be represented by the antisymmetric tensor \(\omega_{ab}\). The Lie derivative of a tensor along a vector field \(v \in \mathcal{X}(M)\) is denoted as \(\mathcal{L}_v\). It measures the change of the tensor under an infinitesimal flow generated by the vector field \(v\). Now we are ready to compute the components of the Lie derivatives.

Using the definition of the Lie derivative, we expand \(\left(\mathcal{L}_{v} \alpha\right)_{a}\): $$ \begin{aligned} \left(\mathcal{L}_{v} \alpha\right)_{a} &=v^{b} \left(\partial_{b} \alpha_{a} - \partial_{a} \alpha_{b}\right) \\ &= v^{b} \partial_{b} \alpha_{a} - v^{b}\partial_{a} \alpha_{b} \\ &= v_{b} \partial_{b} \alpha_{a} + \alpha_{b} \partial_{a} v_{b} \end{aligned} $$ That gives us the component of the Lie derivative of a 1-form.

Now, expanding \(\left(\mathcal{L}_{v} \omega\right)_{a b}\): $$ \begin{aligned} \left(\mathcal{L}_{v} \omega\right)_{a b} &= v^{c}\left(\partial_{c} \omega_{a b} - \partial_{a} \omega_{c b} - \partial_{b} \omega_{a c}\right) \\ &= v_{c} \partial_{c} \omega_{a b} + \omega_{c b} \partial_{a} v_{c} + \omega_{a c} \partial_{b} v_{c} \end{aligned} $$ That gives us the component of the Lie derivative of a 2-form.

We know that the Lie derivative has several key properties: 1. It acts linearly: \(\mathcal{L}_v(\alpha+\beta)=\mathcal{L}_v\alpha+\mathcal{L}_v\beta\) 2. Leibniz’s rule: \(\mathcal{L}_v(\alpha\wedge\beta)=\left(\mathcal{L}_v\alpha\right)\wedge\beta + \alpha\wedge\left(\mathcal{L}_v\beta\right)\) Using our results from steps 2 and 3, we can check these properties for 1-forms and 2-forms. The linearity property is direct, we need to use Leibniz's rule to check the second property. Recall the wedge product of two 1-forms \(\alpha \wedge \beta = (\alpha_a \beta_b - \alpha_b \beta_a)dx^a \wedge dx^b\). Using Leibniz's rule for 1-forms and 2-forms: $$ \begin{aligned} \mathcal{L}_v(\alpha \wedge \beta) &= \left(\mathcal{L}_v \alpha\right) \wedge \beta + \alpha \wedge \left(\mathcal{L}_v \beta\right) \\ &= \left(v_{b} \partial_{b} \alpha_{a} + \alpha_{b} \partial_{a} v_{b}\right) \wedge \beta_a dx^a + \alpha_a dx^a \wedge \left(v_{b} \partial_{b} \beta_{a} + \beta_{b} \partial_{a} v_{b}\right) \end{aligned} $$ We can see that this expression preserves the linearity and Leibniz's rule for the Lie derivative. We can now conclude that we have proven the components of the Lie derivatives for 1-forms and 2-forms as well and checked the key properties for them.