# Definition When the [[determinant]] is thought of as a [[group homomorphism]] between [[Lie Group|Lie groups]] $ \begin{align} \det: GL(n, \mathbb{C}) &\rightarrow \mathbb{C}^*\\ A &\mapsto \det(A) \end{align} $ then the [[trace]] is the [[Induced Lie Algebra Homomorphism]] (since the $\det$ homomorphism above is [[Continuous Map|continuous]]) $ \begin{align} \text{tr}: \mathfrak{gl}(n, \mathbb{C}) &\rightarrow \mathbb{C}\\ X &\mapsto \text{tr}(X) \end{align} $ # Proof Starting from the definition of the induce Lie algebra homomorphism: $ \phi(X) = \frac{d}{dt} \det\left(e^{tX}\right)\bigg|_{t=0} $ and expanding the determinant as: $ \det\left(e^{tX}\right) \approx\det(I + tX) = \epsilon(\pmb{I}_1 + t\pmb{X}_1, \pmb{I}_2 + t\pmb{X}_2, \ldots, \pmb{I}_n + t\pmb{X}_n) $ where the bolded symbols indicated columns of the respective matrices, as per the definition of the determinant, and $\epsilon$ is the [[Levi-Civita tensor]]. Now using the multinealirty of [[tensor|tensors]]: $ \det(e^{tX}) \approx \epsilon\left(\pmb{I_{1}}, \ldots, \pmb{I_{n}}\right)+\epsilon\left(t \pmb{X_{1}}, \pmb{I_{2}}, \ldots, \pmb{I_{n}}\right)+\epsilon\left(\pmb{I_{1}}, t \pmb{X_{2}}, \ldots, \pmb{I_{n}}\right)+\ldots+\epsilon\left(\pmb{I_{1}}, \ldots, \pmb{I_{n-1}}, t \pmb{X_{n}}\right) $ where we have kept only the first order terms in $t$. Then we have: $ \begin{align} \det e^{tX} &\approx \operatorname{det}(I)+\operatorname{det}\left(\begin{array}{ccccc} t X_{11} & 0 & 0 & \ldots & 0 \\ t X_{21} & 1 & 0 & \ldots & 0 \\ t X_{31} & 0 & 1 & \ldots & 0 \\ \vdots & & & \ddots & \vdots \\ t X_{n 1} & 0 & 0 & \ldots & 1 \end{array}\right)+\operatorname{det}\left(\begin{array}{ccccc} 1 & t X_{12} & 0 & \ldots & 0 \\ 0 & t X_{22} & 0 & \ldots & 0 \\ 0 & t X_{32} & 1 & \ldots & 0 \\ \vdots & & & \ddots & \vdots \\ 0 & t X_{n 2} & 0 & \ldots & 1 \end{array}\right)+\ldots+\operatorname{det}\left(\begin{array}{ccccc} 1 & 0 & 0 & \ldots & t X_{1 n} \\ 0 & 1 & 0 & \ldots & t X_{2 n} \\ 0 & 0 & 1 & \ldots & t X_{3 n} \\ \vdots & & & \ddots & \vdots \\ 0 & 0 & 0 & \ldots & t X_{n n} \end{array}\right) .\\ &= 1 + tX_{11} + tX_{22} + \ldots + tX_{nn}\\ &= 1 + t \, \text{tr}(X) \end{align} $ So finally we have: $ \phi(X) = \frac{d}{dt} \left(1 + t\,\text{tr}(X)\right)\bigg|_{t=0} = \text{tr}(X) $ Thus, we have proved that the [[Induced Lie Algebra Homomorphism]] corresponding to: $ \begin{align} \det: GL(n, \mathbb{C}) &\rightarrow \mathbb{C}^*\\ A &\mapsto \det(A) \end{align} $ is $ \begin{align} \text{tr}: \mathfrak{gl}(n, \mathbb{C}) &\rightarrow \mathbb{C}\\ X &\mapsto \text{tr}(X) \end{align} $ # Trace and Determinant Identity Recall that [[Induced Lie Algebra Homomorphism]] have the identity: $ \Phi(e^{tX}) = e^{t\phi(X)} $ applying it to this case with $t=1$ we get: $ \det({e^{X}}) = e^{\text{tr}(X)}, \quad \forall X \in \mathfrak{gl}(n, \mathbb{C}), $ the well known identity.