(Physicist version of) Taylor expansions

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SUMMARY

The discussion focuses on Taylor expansions in the context of Lie algebra elements, specifically for an n-dimensional Lie group with parameters α = (α₁ ... αₙ). The representation D(α) can be expanded around the identity element e, leading to the expression D(dα) = 𝕀 + i dαᵢ Xⁱ, where Xⁱ = -i ∂D(α)/∂αᵢ|_{α=0}. The importance of taking the derivative at α=0 is emphasized, as it defines the point of expansion. Additionally, the limit of (1+a)ⁱ as i approaches infinity is clarified, highlighting the dependence of a on i.

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  • Understanding of Taylor series and polynomial expansions
  • Familiarity with Lie groups and Lie algebras
  • Knowledge of quantum mechanics principles
  • Basic calculus, particularly limits and derivatives
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  • Study Taylor series and their applications in physics
  • Explore the relationship between Lie groups and Lie algebras
  • Learn about the representation theory of Lie groups
  • Investigate limits and convergence in calculus, particularly in relation to Taylor expansions
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Students and professionals in theoretical physics, particularly those focusing on quantum mechanics, representation theory, and mathematical physics. This discussion is beneficial for anyone seeking to deepen their understanding of Taylor expansions in the context of Lie algebra elements.

Jason Bennett
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Homework Statement
in the context of garnering Lie algebras from Lie groups
Relevant Equations
see below
3) Taylor expansion question in the context of Lie algebra elements:

Consider some n-dimensional Lie group whose elements depend on a set of parameters \alpha =(\alpha_1 ... \alpha_n) such that g(0) = e with e as the identity, and that had a d-dimensional representation D(\alpha)=D(g(<br /> \alpha), such that D(0)=\mathbb{1}_{d \times d}. Then in some small neighborhood of \mathbb{1}, we can expand D(\alpha) as,
D(d\alpha) = \mathbb{1} + i d \alpha_i X^i, where X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}

I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:

Please see image [1] below.

Using this terminology on the Lie case:

<br /> \begin{eqnarray}<br /> D(0+d\alpha) &amp;=&amp; D(0) + \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha<br /> \\<br /> &amp;=&amp; \mathbb{1} + (i) (-i) \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha<br /> \\<br /> &amp;=&amp; \mathbb{1} + (i) X^i d\alpha<br /> \end{eqnarray}<br />

is this correct? Also, why is the "taking the derivative at \alpha=0 important? And can you please point me towards a place to learn these types of Taylor expansions?

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of (1+a)^x as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
[1]: https://i.stack.imgur.com/yAXum.png
 
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Jason Bennett said:
Homework Statement: in the context of garnering Lie algebras from Lie groups
Homework Equations: see below

3) Taylor expansion question in the context of Lie algebra elements:

Consider some n-dimensional Lie group whose elements depend on a set of parameters \alpha =(\alpha_1 ... \alpha_n) such that g(0) = e with e as the identity, and that had a d-dimensional representation D(\alpha)=D(g(<br /> \alpha), such that D(0)=\mathbb{1}_{d \times d}. Then in some small neighborhood of \mathbb{1}, we can expand D(\alpha) as,
D(d\alpha) = \mathbb{1} + i d \alpha_i X^i, where X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}

I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:

Please see image [1] below.

Using this terminology on the Lie case:

<br /> \begin{eqnarray}<br /> D(0+d\alpha) &amp;=&amp; D(0) + \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha<br /> \\<br /> &amp;=&amp; \mathbb{1} + (i) (-i) \frac{<br /> \partial D(\alpha)}{\partial \alpha_i}d\alpha<br /> \\<br /> &amp;=&amp; \mathbb{1} + (i) X^i d\alpha<br /> \end{eqnarray}<br />

is this correct? Also, why is the "taking the derivative at \alpha=0 important? And can you please point me towards a place to learn these types of Taylor expansions?

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of (1+a)^x as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
[1]: https://i.stack.imgur.com/yAXum.png

Your question is not specific to Lie algebras/groups. You simply don't know Taylor series.

A Taylor series is a polynomial expansion of a function around a certain point:

##f(x) =\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}##

##a## is the point around which you are expanding.

Since you wrote ##
D(0+d\alpha) = D(0) + \frac{\partial D(\alpha)}{\partial \alpha_i}d\alpha##, the point you are expanding around is ##\alpha = 0##, since the right hand side has the constant term ##D(0)##. That's why the derivative is at zero.

In equations 1,2,3, ##d\alpha## should be changed to ##d\alpha _i## and there should be a sum over i.

As for the limit of ##(1+a)^x##, you are forgetting that a depends on x. When ##a = c/x##, you get ##e^c## as ##x \to \infty##.
 
Jason Bennett said:
Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of (1+a)^x as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
Look at 9.13 a bit more carefully. It's not the limit of ##(1+a)^N## as ##N \to \infty##; it's the limit of ##(1+\delta a)^N## where ##\delta a = a/N##.
 

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