3
$\begingroup$

I am asked to show that if $\mathfrak{g}$ is a semisimple Lie Algebra with root system of type $G_2$, then it has a unique, $7$-dimensional faithful representation.

To start, let $\omega_1, \omega_2$ be the fundamental weights of the root system $G_2$ and suppose that $\lambda = m_1\omega_1 + m_2\omega_2$ is a dominant weight. Then by using the Weyl Dimension Formula, we see that:

$$\dim(V(\lambda)) = \frac{1}{120}(m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1 + 3m_2+5).$$

Using this formula we see that $V(\omega_1)$ is the only irreducible representation of $\mathfrak{g}$ and there are no irreducible representations of dimensions $2$, $3$, $4$, $5$, or $6$. This in turn means there are no representations of dimensions $2$, $3$, $4$, $5$, or $6$ not of the form $V(0) ^{\otimes n}$

Finally, we note $V(0)^{\otimes 7}$ is not faithful, and so we must be required to show that the representation $V(\omega_1)$ is faithful.

It is here that I am stuck since I do not know how to do this.

My immediate thought is that if we assume we do not have injectivity then we can find a subrepresentation, though I am unsure of how to do this, or how to use the fact that the representation is specifically $V(\omega_1)$ and not just any $7$-dimensional vector space. Below is my attempt

Let $\phi$ be the representation $\phi : \mathfrak{g} \rightarrow \mathfrak{gl}(V(\omega_1))$ and suppose that for some $x \neq y, x,y \in \mathfrak{g}$ we have that $\phi(x) = \phi(y)$

If $x = ky$ for some $k\in \mathbb C$ then by linearity of $\phi$, we have $\phi(x) = 0$

Let $v_1, \dotsc, v_n$ be a basis for $\mathfrak{g}$ with $v_1 = x$. Then as $\phi$ is a homomorphism, we must have that $\dim(\phi(\mathfrak{g})) < \dim(\mathfrak{g})$.

[I am unable to pursue this approach any further].

In the case that $x$ and $y$ are linearly independent, then I have even fewer ideas.

I think ultimately, treating $\mathfrak{gl}(V(\omega_1))$ as a vector space of matrices, then I would like to show that in $\phi(\mathfrak{g})$ there is an entire row or column that is not in use. Therefore, we may ignore the corresponding basis vector(s) of $V(\omega_1)$ to get a subrepresentation and thus a contradiction.

However, I am unsure of how to achieve this and would appreciate any help.

$\endgroup$
2
  • 2
    $\begingroup$ Just use the fact that your Lie algebra is simple, so the kernel of your representation is either zero and thus the representation is faithful, or the kernel is everything and the representation is thus trivial $\endgroup$
    – user8268
    Jan 9, 2019 at 20:14
  • $\begingroup$ @user8268 Thank you! I seem to have over complicated the issue and forgotten then $G_2$ connected implies $\Phi$ is irreducible which implies $\mathfrak{g}$ is simple. $\endgroup$
    – user366818
    Jan 9, 2019 at 20:48

1 Answer 1

2
$\begingroup$

It is clear that the simple module of smallest dimension, different from the trivial one, is $L(\omega_1)$, which has dimension $7$. Since ${\mathfrak g}_2$ is a simple Lie algebra, the module is faithful. Every ${\mathfrak g}_2$-module is semisimple, i.e., direct sum of simple modules.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .