2 added 27 characters in body

Here is a proof of the Weyl dimension formula, up to a constant factor, that does not go through the Weyl character formula. I leave up to you how geometric you consider this.

Set $\mu=\lambda+\rho$. It is clear that the answer will be a polynomial in $\mu$, of degree $\leq \dim_{\mathbb{C}} X$.

Lemma: This polynomial is $W$-antisymmetric.

Proof: To see this, it will be easier to work with the description of $X$ as $K/T$, where $K$ is the compact real form of $G$, and $T$ the maximal torus. There are two $W$ actions on $K/T$: the one from the left, which exists because $K/T$ is a coset space, and the one from the right, which exists because $W$ normalizes $T$. We will be concerned with the second action. Note that this action is not holomorphic, and is very hard to see from the $G/B$ description of $X$.

Under the identification of $H^2(X)$ with $\mathfrak{t}^*$, this $W$ action induces the standard reflection representation of $W$. On top cohomology, $W$ acts by the sign character. Using the latter fact, for any differential form $\eta$, we have $\int_X w^*(\eta) = (-1)^{\ell(w)} \int_X \eta$. Let's apply this to the integrand we care about.

We write $\Phi^{+}$ for the set of positive roots. $$w^* \left( e^{\mu} \prod_{\alpha \in \Phi^{+}} \frac{\alpha}{e^{\alpha/2} - e^{- \alpha/2}} \right) = e^{w(\mu)} \prod_{\alpha \in w \Phi^{+}} \frac{\alpha}{e^{\alpha/2} - e^{- \alpha/2}}$$ We can partition $w \Phi^{+}$ into $A := \Phi^{+} \cap w \Phi^{+}$ and $B = \Phi^{-} \cap w \Phi^{+}$. Then $A \sqcup (-B) = \Phi^{+}$. Abbreviate $T(\alpha) := \alpha/(e^{\alpha/2} - e^{- \alpha/2})$ and note that $T(\alpha) = T(- \alpha)$. So $$\prod_{\alpha \in w \Phi^{+}} T(\alpha) = \prod_{\alpha \in A } T(\alpha) \prod_{\alpha \in B} T(\alpha) = \prod_{\alpha \in A } T(\alpha) \prod_{\alpha \in B} T(-\alpha)= \prod_{\alpha \in \Phi^{+}} T(\alpha)$$

Putting it all together, we see that $$\int_X e^{w \mu} \prod_{\alpha \in \Phi^{+}} T(\alpha) = \int_X w^* \left( e^{\mu} \prod_{\alpha \in \Phi^{+}} T(\alpha) \right) = (-1)^{\ell(w)} \int_X e^{\mu} \prod_{\alpha \in \Phi^{+}} T(\alpha)$$

This proves the lemma. $\square$

Now, let $\beta$ be any root in $\Phi^{+}$. Suppose that $\langle \mu, \beta \rangle =0$. Let $t$ be the reflection which negates $\beta$. Then $t(\mu)=\mu$, and $(-1)^{\ell(t)} = -1$, so our polynomial must vanish at $\mu$. Thus, our polynomial is divisible by $\langle \cdot , \beta \rangle$. Since our polynomial has degree $\leq \dim_{\mathbb{C}} X = |\Phi^{+}|$, it must be $c \prod_{\beta \in \Phi^{+}} \langle \cdot , \beta \rangle$ for some constant $c$. I found a couple of ways to work out the constant, but none of them were as clean as I'd like, so I'll leave it there.

1

Here is a proof of the Weyl dimension formula, up to a constant factor, that does not go through the Weyl character formula. I leave up to you how geometric you consider this.

Set $\mu=\lambda+\rho$. It is clear that the answer will be a polynomial in $\mu$, of degree $\leq \dim_{\mathbb{C}} X$.

Lemma: This polynomial is $W$-antisymmetric.

Proof: To see this, it will be easier to work with the description of $X$ as $K/T$, where $K$ is the compact real form of $G$, and $T$ the maximal torus. There are two $W$ actions on $K/T$: the one from the left, which exists because $K/T$ is a coset space, and the one from the right, which exists because $W$ normalizes $T$. We will be concerned with the second action. Note that this action is not holomorphic, and is very hard to see from the $G/B$ description of $X$.

Under the identification of $H^2(X)$ with $\mathfrak{t}^*$, this $W$ action induces the standard reflection representation of $W$. On top cohomology, $W$ acts by the sign character. Using the latter fact, for any differential form $\eta$, we have $\int_X w^*(\eta) = (-1)^{\ell(w)} \int_X \eta$. Let's apply this to the integrand we care about.

We write $\Phi^{+}$ for the set of positive roots. $$w^* \left( e^{\mu} \prod_{\alpha \in \Phi^{+}} \frac{\alpha}{e^{\alpha/2} - e^{- \alpha/2}} \right) = e^{w(\mu)} \prod_{\alpha \in w \Phi^{+}} \frac{\alpha}{e^{\alpha/2} - e^{- \alpha/2}}$$ We can partition $w \Phi^{+}$ into $A := \Phi^{+} \cap w \Phi^{+}$ and $B = \Phi^{-} \cap w \Phi^{+}$. Then $A \sqcup (-B) = \Phi^{+}$. Abbreviate $T(\alpha) := \alpha/(e^{\alpha/2} - e^{- \alpha/2})$ and note that $T(\alpha) = T(- \alpha)$. So $$\prod_{\alpha \in w \Phi^{+}} T(\alpha) = \prod_{\alpha \in A } T(\alpha) \prod_{\alpha \in B} T(\alpha) = \prod_{\alpha \in A } T(\alpha) \prod_{\alpha \in B} T(-\alpha)= \prod_{\alpha \in \Phi^{+}} T(\alpha)$$

Putting it all together, we see that $$\int_X e^{w \mu} \prod_{\alpha \in \Phi^{+}} T(\alpha) = \int_X w^* \left( e^{\mu} \prod_{\alpha \in \Phi^{+}} T(\alpha) \right) = (-1)^{\ell(w)} \int_X e^{\mu} \prod_{\alpha \in \Phi^{+}} T(\alpha)$$

This proves the lemma. $\square$

Now, let $\beta$ be any root in $\Phi^{+}$. Suppose that $\langle \mu, \beta \rangle =0$. Let $t$ be the reflection which negates $\beta$. Then $t(\mu)=\mu$, so our polynomial must vanish at $\mu$. Thus, our polynomial is divisible by $\langle \cdot , \beta \rangle$. Since our polynomial has degree $\leq \dim_{\mathbb{C}} X = |\Phi^{+}|$, it must be $c \prod_{\beta \in \Phi^{+}} \langle \cdot , \beta \rangle$ for some constant $c$. I found a couple of ways to work out the constant, but none of them were as clean as I'd like, so I'll leave it there.